Math Problems for 8 – 15 – 16

Problem 1

For which function defined by a polynomial are the zeros of the polynomial −4 and −6?

1)  y = x2 − 10x − 24

2)  y = x2 + 10x + 24

3)  y = x2 + 10x − 24

4)  y = x2 − 10x + 24

Discussion and Solution:

There are two ways of solving this problem – (1) solve each equation for the value of x and (2) substitute each possible solution into each equation to which one works. Let’s try them both.

Method (1) –

  • y = x2 – 10x – 24

0 = x2 – 10x – 24

0 = (x – 4)(x – 6)

0 = (x – 4); x = 4

0 = (x – 6); x = 6

No way

  • y = x2 + 10x + 24

0 = x2 + 10x + 24

0 = (x + 4)(x + 6)

0 = (x + 4); x = – 4

0 = (x + 6); x = – 6

Bingo – they work here – we can stop here since we have the answer

Method (2) –

  • y = x2 – 10x – 24

0 = x2 – 10x – 24

0 = (– 4)2 – 10(– 4) – 24

0 = 16 + 40 – 24

0 ≠ 32

No way

  • y = x2 + 10x + 24

0 = x2 + 10x + 24

0 = (– 4)2 + 10(– 4) + 24

0 = 16 – 40 + 24

0 = 0

This one works, but we better check the other one.

y = x2 + 10x + 24

0 = x2 + 10x + 24

0 = (– 6)2 + 10(– 6) + 24

0 = 36 – 60 + 24

Bingo. This one works also.

Correct solution is (2).

Problem 2

If f(x) = x2 − 2x − 8 and g(x) = ¼ x − 1, for which values of x is f(x) = g(x)?

1) −1.75 and −1.438

2) −1.75 and 4

3) −1.438 and 0

4) 4 and 0

Discussion and Solution:

The easiest way to solve this is to substitute each possible solution into each equation and see which one clicks.

Since options (3) and (4) both have 0 as a possible solution, let’s start with this one and see what happens.

f(x) = x2 − 2x – 8 = (0)2 – 2(0) – 8 = – 8

g(x) = ¼ x – 1 = ¼ (0) – 1 = – 1

I don’t know about you, but I was taught – 8 ≠ – 1. Therefore, choices (3) and (4) are out.

Now let’s try options (1) and (2). Let’s try the 4.

f(x) = x2 − 2x – 8 = (4)2 – 2(4) – 8 = 16 – 8 – 8 = 0

g(x) = ¼ x – 1 = ¼ (4) – 1 = 1 – 1 = 0

 That works, now let’s try the – 1.75.

f(x) = x2 − 2x – 8 = (– 1.75)2 – 2(– 1.75) – 8 = 3.0625 + 3.5 – 8 = – 1.4375

g(x) = ¼ x – 1 = ¼ (– 1.75) – 1 = – 0.4375 – 1 = – 1.4375

That’s okay, but let’s try the – 1.438 to rule out choice (1).

f(x) = x2 − 2x – 8 = (– 1.438)2 – 2(– 1.438) – 8 = 2.0678 + 2.876 – 8 = – 3.056

g(x) = ¼ x – 1 = ¼ (– 1.438) – 1 = – 0.3595 – 1 = – 1.3595

Nope.

Choice (2) is the correct choice.

Problem 3

Which system of equations has the same solution as the system below?

2x + 2y = 16

3x y = 4

1)  2x + 2y = 16

6x − 2y = 4

2)  2x + 2y = 16

6x − 2y = 8

3)  x + y = 16

3x y = 4

4)  6x + 6y = 48

6x + 2y = 8

 Discussion and Solution:

In order to have the same solution, when we divide the possible solutions by some small number, say a 2 or 3, we should arrive at the original equation. Here goes.

Choice (1):

1)  2x + 2y = 16

6x − 2y = 4

The first equation is the same as one of the original equations. As for the second equation, the left side appears to be doubled from the original equation, but not the right side. This choice is out.

Choice (2):

2)  2x + 2y = 16

6x − 2y = 8

The first equation is the same as one of the original equations. As for the second equation, both sides have been doubled as compared to the other original equation. This choice is the correct one.

Choice (3):

3)  x + y = 16

3x y = 4

In this choice, the second equation is the same as one of the original equations, but the first equation is messed up. The right side of this equation is the same as the right side of the other original equation, but the left side has been halved. This choice is out.

Choice (4):

4)  6x + 6y = 48

6x + 2y = 8

In this choice, the first equation is the triple of one of the original equations, but the second one is messed up. At first glance, it appears that the equation has merely been doubled. Take a closer look. If it were just doubled, it would be 6x – 2y = 8, not 6x + 2y = 8. This choice is out.

Choice (2) is the correct one.

Problem 4

Mo’s farm stand sold a total of 165 pounds of apples and peaches. She sold apples for $1.75 per pound and peaches for $2.50 per pound. If she made $337.50, how many pounds of peaches did she sell?

1)  11

2)  18

3)  65

4)  100

 Discussion and Solution:

Let’s set up two equations and solve. a = number pounds of apples. p = number of pounds of peaches.

a + p = 165

1.75a + 2.50p = 337.50

Two equations with two unknowns. Don’t you just love that? As you’re probably aware, when we have two equations with two unknowns, there are three ways of solving this. They are:

(a) Solve one equation for one of the unknowns and substitute this into the other equation and solve.

(b) Graph the two equations and look to see where they intersect. This point of intersection will indicate the solutions. If the lines are parallel or on top of each other, then there is no common solution.

(c) Multiply one of the equations by some number (in this case, multiply the top equation by 1.75) and subtract the two equations, thereby eliminating one of the variables. Solve for the remaining variable. Now, substitute this value into either of the original equations and determine the missing value.

Let’s use procedure (a).

a + p = 165

a = 165 – p

 

1.75a + 2.50p = 337.50

1.75(165 – p) + 2.50p = 337.50

288.75 – 1.75p + 2.50p = 337.50

288.75 + 0.75p = 337.50

0.75p = 337.50 – 288.75 = 48.75

p = 65

Choice (3) is the correct choice.

Problem 5

If A = 3x2 + 5x − 6 and B = −2x2 − 6x + 7, then A B equals

1) −5x2 − 11x + 13

2) 5x2 + 11x − 13

3) −5x2x + 1

4) 5x2x + 1

 Discussion and Solution:

Since we’re looking for AB, we merely need to set the two equations equal to each other.

AB

(3x2 + 5x – 6) – (−2x2 − 6x + 7)

3x2 + 5x – 6 + 2x2 − 6x – 7

5x2 + 11x – 13

Choice (2) is correct.

Physics Problems for 8 – 12 – 16

Problem 1

Which quantity is scalar?
(1) mass
(2) force
(3) momentum
(4) acceleration

¿Qué cantidad es escalar?

(1) masa
(2) fuerza
(3) momento
(4) aceleración

Discussion and Solution

A scalar quantity, as opposed to a vector quantity, has only a magnitude (numerical) component, but no direction component. Choice (1)

Problem 2

What is the final speed of an object that starts from rest and accelerates uniformly at 4.0 meters per second2 over a distance of 8.0 meters?
(1) 8.0 m/s
(2) 16 m/s
(3) 32 m/s
(4) 64 m/s

¿Cuál es la velocidad final de un objeto que comienza desde descanso y acelera uniformemente a 4.0 metros por segundo2 sobre una distancia de 8.0 metros?
(1) 8.0 m/s
(2) 16 m/s
(3) 32 m/s
(4) 64 m/s

Discussion and Solution

Vf2 = Vi2 + 2AD, where Vf = final velocity, Vi = initial velocity, A = acceleration, and D = distance.
Vf2 = Vi2 + 2AD = (0.0 m/s)2 + 2(4.0 m/s2)(8.0 m) = 64 m2/s2
Taking the square root of both sides, we get Vf = 8.0 m/s, Choice (1).

Problem 3

The components of a 15-meters-per-second velocity at an angle of 60.° above the horizontal are
(1) 7.5 m/s vertical and 13 m/s horizontal
(2) 13 m/s vertical and 7.5 m/s horizontal
(3) 6.0 m/s vertical and 9.0 m/s horizontal
(4) 9.0 m/s vertical and 6.0 m/s horizontal

Los componentes de una velocidad de 15 metros por segundo a un ángulo de 60° sobre la horizontal son
(1) 7.5 m/s vertical y 13 m/s horizontal

(2) 13 m/s vertical y 7.5 m/s horizontal

(3) 6.0 m/s vertical y 9.0 m/s horizontal

(4) 9.0 m/s vertical y 6.0 m/s horizontal

Discussion and Solution

Think of a Right Triangle. Think of the definition of Sin and Cos. Sin = Opposite / Hypotenuse. Cos = Adjacent / Hypotenuse.
Sin 60o = X / 15 m/s. X = (15 m/s)Sin60o = 13 m/s vertical.
Cos 60o = X / 15 m/s. X = (15 m/s)Cos 60o = 7.5 m/s horizontal
Choice (2).

Problem 4

What is the time required for an object starting from rest to fall freely 500. meters near Earth’s surface?
(1) 51.0 s
(2) 25.5 s
(3) 10.1 s
(4) 7.14 s

¿Cuál es el tiempo requerido para que un objeto que comienza desde descanso caiga libremente 500 metros cerca de la superficie de la Tierra?
(1) 51.0 s
(2) 25.5 s
(3) 10.1 s
(4) 7.14 s

Discussion and Solution

D = ViT + ½AT2
Since we are starting from rest, Vi = 0.0 m/s. This alters our equation to: D = ½AT2. Solving for T, we get T = SQRT(2D/A) = SQRT(2[500. m] / 9.8 m/s2) = 10.1 s. Choice (3).

Problem 5

A baseball bat exerts a force of magnitude F on a ball. If the mass of the bat is three times the mass of the ball, the magnitude of the force of the ball on the bat is
(1) F
(2) 2F
(3) 3F
(4) F/3

Un bate de béisbol ejerce una fuerza de magnitud F en una bola. Si la masa del bate es tres veces la masa de la bola, la magnitud de la fuerza de la bola en el bate es
(1) F
(2) 2F
(3) 3F
(4) F/3

Discussion and Solution

Remember, a Force exerted on an object one way, is counter balanced by a Force of equal magnitude exerted in the opposite direction. Choice (1).

Chemistry Problems for 8 – 10 – 16

 

Problem 1

Which particles have approximately the same mass?

(1)       alpha particle and beta particle

(2)       alpha particle and proton

(3)       neutron and positron

(4)       neutron and proton

 

¿Cuáles partículas tienen aproximadamente la misma masa?

(1)        La partícula alfa y la partícula beta

(2)        La partícula alfa y protón

(3)        neutrón y positrón

(4)        neutrón y protón

 

Solution and Discussion

To answer this question depends upon your remembering the masses of the various nuclear particles:

Alpha particle = 4

Beta particle = 0

Neutron = 1

Positron = 0

Proton = 1

From the above, a neutron and a proton both have a mass of 1, therefor Choice (4) is the correct choice.

 

Problem 2

Which phrase describes an atom?

(1)       a negatively charged nucleus surrounded by positively charged protons

(2)       a negatively charged nucleus surrounded by positively charged electrons

(3)       a positively charged nucleus surrounded by negatively charged protons

(4)       a positively charged nucleus surrounded by negatively charged electrons

 

¿Qué frase describe un átomo?

(1)        un núcleo cargado negativamente rodeado por protones cargados positivamente

(2)        un núcleo cargado negativamente rodeado por electrones cargados positivamente

(3)        un núcleo cargado positivamente rodeado por protones cargados negativamente

(4)        un núcleo cargado positivamente rodeado por electrones cargados negativamente

 

Solution and Discussion

To answer this problem, you need to remember what an atom consists of.

The nucleus has a positive charge due to the presence of positively charged protons.

The nucleus is surrounded by electrons. Electrons are negatively charged.

Therefor Choice (4) is the correct choice.

 

Problem 3

An orbital is defined as a region of the most probable location of

(1)       an electron

(2)       a neutron

(3)       nucleus

(4)       proton

 

Un orbital se define como una región de la más probable localización de

(1)        un electrón

(2)        un neutrón

(3)        un núcleo

(4)        un protón

 

Solution and Discussion

To answer this problem, you need to remember the definition of an orbital.

An orbital is defined as a region of the most probable location of electrons.

Choice (1) is the correct choice.

 

Problem 4

The bright-line spectrum of an element in the gaseous phase is produced as

(1)       protons move from lower energy states to higher energy states

(2)       protons move from higher energy states to lower energy states

(3)       electrons move from lower energy states to higher energy states

(4)       electrons move from higher energy states to lower energy states

 

El espectro de líneas brillantes de un elemento en la fase gaseosa se produce cuando

(1)        los protones pasan de estados de menor energía a estados de mayor energía

(2)        los protones pasan de estados de mayor energía a estados de energía más bajos

(3)        los electrones se mueven a partir de estados de menor energía a estados de mayor energía

(4)        los electrones se mueven a partir de los estados de mayor energía a estados de energía más bajos

 

Solution and Discussion

The answer to this problem is dependent upon knowing what produces bright-line spectrum.

The bright-line spectrum of an element in the gaseous phase is produced as electrons move from higher energy states to lower energy states. You probably did a lab to show this.

Choice (4) is the correct choice.

 

Problem 5

An atom of lithium-7 has an equal number of

(1)       electrons and neutrons

(2)       electrons and protons

(3)       positrons and neutrons

(4)       positrons and protons

 

Un átomo de litio-7 tiene un número igual de

(1)        electrones y neutrones

(2)        electrones y protones

(3)        positrones y neutrones

(4)        positrones y protones

 

Solution and Discussion

All atoms, being electrically neutral, contain an equal number of positive charges (protons) and negative charges (electrons), Choice (2).

Math Problems for 8 – 8 – 16

Problem 1

Factored completely, the expression 6x x^3 – x^2 is equivalent to

(1)  x(x +3)(x −2)

(2)  x(x −3)(x +2)

(3)  −x(x −3)(x +2)

(4)  −x(x +3)(x −2)

 

Solution and Discussion

Looking at the equation, every term contains at least one x, therefore we can factor it out, getting

x(6 – x^2 – x).

Next we would rearrange the equation to get it into normal form.

x(-x^2 – x + 6)

This looks odd since we are starting with a term that contains a negative sign. Perhaps we had better factor that out also, getting

x(x^2 + x – 6)

Now we can factor the parenthetical expression, getting

x(x + 3)(x – 2)

 

Problem 2

Factored completely, the expression 12x^4  + 10x^3  −12x^2 is equivalent to

(1)  x^2 (4x +6)(3x −2)

(2)  2(2x^2 +3x)(3x^2 −2x)

(3)  2x^2(2x −3)(3x +2)

(4)  2x^2(2x +3)(3x −2)

 

Solution and Discussion

It looks like every term in the expression contains at least x^2, so let’s factor that out.

x^2(12x^2 + 10x – 12)

Now let’s factor out a 2, since every term is divisible by 2.

2x^2(6x^2 + 5x – 6)

The final step is a little tricky. Sorry. The first term, 6, factors into a 3 and a 2, so we know that the beginning of the final answer looks like this.

2x^2(3x       )(2x       )

The final term, also a 6, factors into a 3 and a 2. Now we don’t know whether to put the 3 with the 3x term or with the 2x term. This also holds for the 2. To solve this dilemma, we have to consider FOIL. The First terms multiply to a 6. How do we get the sum of the Outer and the Inner terms add to get a 5? Try this:

2x^2(3x – 2)(2x + 3)

 

Problem 3

Factor completely: 10ax^2   − 23ax  − 5a

 

Solution and Discussion

Every term contains an a, so let’s factor it out, getting

a(10x^2 – 23x – 5)

If we look at the 10, it factors into a 5 and a 2, so we know this:

a(5x       )(2x      )

The last term in the expression, the 5, factors into a 1 and a 5. The problem is which of these goes with the 5x and which goes with the 2x? To answer this, we have to remember FOIL, particularly the sum of the Outer and Inner terms. How about this?

a(5x + 1)(2x – 5)

 

Problem 4

Factor the expression 12t^8   − 75t^4 completely.

 

Solution and Discussion

Each term contains a t^4, so let’s factor it out, getting

t^4(12t^2 – 75)

Now we also factor out a 3, getting

3t^4(4t^2 – 25)

This looks like the difference of squares. Something like a^2 – b^2 =

(a – b)(a+2). Putting this thought into our problem, we get

3t^4(2t – 5)(2t + 5)

 

Problem 5

When factored completely, x^3  + 3x^2  − 4x  – 12 equals

(1)  (x +2)(x −2)(x −3)

(2)  (x +2)(x −2)(x +3)

(3)  (x^2 −4)(x +3)

(4)  (x^2 −4)(x −3)

 

Solution and Discussion

First we group the first two terms together, getting

(x^3 + 3x^2) – 4x – 12

Grouping together the last two terms we get

(x^3 + 3x^2) – (4x + 12)

Looking at the first group, we factor out x^2 and for the last group, we factor out a 4. Doing this, we get

x^2(x + 3) – 4(x + 3)

Since both terms contain (x + 3), we can factor this out of the overall equation, getting

(x^2 – 4)(x + 3)

Looking at the first term, it looks like the difference of squares.

(x^2 – 4) = (x + 2)(x – 2). Putting this into the equation, we get

(x + 2)(x – 2)(x + 3)

Choice (2).

Physics Problems for 8 – 5 – 16

Problem 1

Which quantities are scalar?

(1) speed and work

(2) velocity and force

(3) distance and acceleration

(4) momentum and power

 

¿Qué cantidades son escalar?

(1) rapidez y trabajo

(2) velocidad y fuerza

(3) distancia y aceleración

(4) momento y potencia

 

Answer and discussion:

Answering this question correctly depends on your knowing the difference between scalar and vector quantities. Scalar quantities have only magnitude (a numerical value), including a unit of measure. On the other hand, vector quantities have both magnitude and direction. Let’s look at the situation for each of the choices and then make a CORRECT decision.

In choice (1), both speed and work have only magnitude. Hence choice (1) is the answer. By the way, don’t confuse speed with velocity (as in choice (2)). Speed has only magnitude, where velocity has both magnitude AND direction, e.g., 5 mi/hr South.

In choice (2), velocity is a vector quantity, as is force. For example, 69 km/hr North and 18 Newtons East, respectively.

In choice (3), distance is a scalar quantity, whereas acceleration is a vector quantity. Distance is measured with only a magnitude (8 mi). Don’t confuse displacement with distance. Displacement would be, for example, 80 mi North. Acceleration is a vector quantity, e.g., 17mi/hr/hr West.

In choice (4), momentum is calculated by multiplying an object’s mass and its velocity. Mass is measured in kg (a scalar quantity). Velocity is a vector quantity. Hence the product of the two would be a vector quantity. Power is determined by dividing work by time, both scalar quantities.

 

Problem 2

A 3.00-kilogram mass is thrown vertically upward with an initial speed of 9.80 meters per second. What is the maximum height this object will reach? [Neglect friction.]

(1) 1.00 m

(2) 4.90 m

(3) 9.80 m

(4) 19.6 m

 

Una masa de 3.00-kilogramo es lanzada verticalmente hacia arriba con una velocidad inicial de 9.80 metros por segundo. ¿Cuál es la altura máxima que este objeto alcanzará? [Omita fricción.]

(1) 1.00 m

(2) 4.90 m

(3) 9.80 m

(4) 19.6 m

 

Answer and discussion:

Specifying that a 3.00-kilogram mass is thrown is a distractor. It doesn’t matter what is thrown upward. The maximum height would be the same.

One would determine the maximum height with the formulas: vf = vi + at  and  d = vit + ½ at^2, where vf = final velocity, vi = initial velocity, t = time, d = distance, and a = acceleration.

The final velocity at the peak, vf, is 0 m/s. The initial velocity, vi, is 9.8 m/s. The rate of acceleration is -9.8 m/s^2. Putting these values into the equation, we get 0 m/s = 9.8 m/s + (9.8 m/s^2)t. Solving for t, we get: t = 1 sec.

Next we use the second equation: d = vit + ½ at^2. Putting in what we know at this point, we get d = (9.8 m/s)(1 s) + ½ (-9.8 m/s^2)(1 s)^2. We now solve for d, getting d = 4.90 m, choice (2).

 

Problem 3

An airplane traveling north at 220. meters per second encounters a 50.0-meters-per-second crosswind from west to east. What is the resultant speed of the plane?

(1) 170. m/s

(2) 214 m/s

(3) 226 m/s

(4) 270. m/s

 

Un avión viajando al norte a 220 metros sobre segundo encuentra un viento transversal de 50.0-metros-sobre-segundo que viene de oeste a este.

 

¿Cuál es la velocidad resultante del avión?

(1) 170. m/s

(2) 214 m/s

(3) 226 m/s

(4) 270. m/s

 

Answer and discussion:

To solve this problem requires remember some of your algebra. What we have in the problem is a right triangle. We are given the values for two of the three sides. We need to use the Pythagorean theorem:  c^2 = a^2 + b^2. We are looking for the hypotenuse, c. Putting what the value of the two sides, we get: c^2 = (50.0 m/s)^2 + (220. m/s)^2 = 2500 m^2/s^2 + 48400 m^2/s^2 = 50900 m^2/s^2. Solving for c, we get: c = √50900 m^2/s^2 = 225.610… m/s = 226 m/s, choice (3).

 

Problem 4

A 160.-kilogram space vehicle is traveling along a straight line at a constant speed of 800. meters per second. The magnitude of the net force on the space vehicle is

(1) 0 N

(2) 1.60 x10^2 N

(3) 8.00 x10^2 N

(4) 1.28 x10^5 N

 

Un vehículo espacial de 160.-kilogramo está viajando a lo largo de una línea recta a una velocidad constante de 800 metros por segundo. La magnitud de la fuerza neta en el vehículo espacial es

(1) 0 N

(3) 8.00 × 10^2 N

(2) 1.60 × 10^2 N

(4) 1.28 × 10^5 N

 

Answer and discussion:

Since the space vehicle is traveling along a straight line at a constant speed, it obeys Newton’s Second Law of Motion. Expressed mathematically F = ma. When we put in what values we know, we get F = (160. kg)(0 m/s^2) = 0 N. (The acceleration is 0 because the vehicle is traveling along a straight line at a constant speed.) Note that we never used the 800. m/s. It is a distractor. It could have been any number and we would still get 0 N, choice (1).

 

Problem 5

A student throws a 5.0-newton ball straight up. What is the net force on the ball at its maximum height?

(1) 0.0 N

(2) 5.0 N, up

(3) 5.0 N, down

(4) 9.8 N, down

 

Un estudiante tira una bola de 5.0-newton directo hacia arriba. ¿Cuál es la fuerza neta en la bola a su altura máxima?

(1) 0.0 N

(2) 5.0 N, arriba

(3) 5.0 N, abajo

(4) 9.8 N, abajo

 

Answer and discussion:

At the ball’s maximum height, there is no longer any acceleration, so the regular formula we would use, F = ma, is not applicable. Therefore, the only force acting on the ball is that caused by gravity. Since the ball weighs 5.0-Newtons, and the force of gravity is downward, choice (3) is correct.

 

 

Chemistry Problems for 8-3-15

Problem 1

Compared to an electron, which particle has a charge that is equal in magnitude but opposite in sign?

(1) an alpha particle

(2) a beta particle

(3) a neutron

(4) a proton

 

Comparado con un electrón, ¿qué partícula tiene una carga que es igual en magnitud pero opuesta en signo?

(1) una partícula alfa

(2) una partícula beta

(3) un neutrón

(4) un protón

 

Answer and discussion:

In choice (1), remember that an alpha particle has a charge of +2 and a mass of 4. In choice (2), a beta particle has a charge of -1 and a mass of 0. In choice (3), a neutron has no charge and a mass of 1. In choice (4), a proton has a charge of +1 and a mass of 1.

Since an electron has a charge of -1, the correct answer is choice (4).

 

Problem 2

The mass of a proton is approximately equal to

(1) 1 atomic mass unit

(2) 12 atomic mass units

(3) the mass of 1 mole of carbon atoms

(4) the mass of 12 moles of electrons

 

La masa de un protón es aproximadamente igual a

(1) 1 unidad de masa atómica

(2) 12 unidades de masa atómica

(3) la masa de 1 mol de átomos de carbono

(4) la masa de 12 moles de átomos de carbono

 

Answer and discussion:

The mass of a proton is 1 by straight memorization of the mass and charges of elementary particles. The 1 meaning 1 atomic mass unit, or amu. Choice (1) is the correct answer.

 

Problem 3

Which property decreases when the elements in Group 17 are considered in order of increasing atomic number?

(1) atomic mass

(2) atomic radius

(3) melting point

(4) electronegativity

 

¿Cuál propiedad disminuye cuando los elementos en el Grupo 17 son considerados en orden creciente de número atómico?

(1) masa atómica

(2) radio atómico

(3) punto de fusión

(4) electronegatividad

 

Answer and discussion:

Electronegativity decreases as the atoms get larger and larger, since the attraction of the outer electrons are under less of the attraction of the protons in the nucleus.

As one gets increasing number of protons in the nucleus (increasing atomic number), the size of the atom gets larger and larger.

Putting these two together, choice (4) is the correct choice.

 

Problem 4

Any substance composed of two or more elements that are chemically combined in a fixed proportion is

(1) an isomer

(2) an isotope

(3) a solution

(4) a compound

 

Cualquier sustancia compuesta de dos o más elementos que están químicamente combinados en una proporción fija es

(1) un isómero

(2) un isotopo

(3) una solución

(4) un compuesto

 

Answer and discussion:

Isomers are two molecules that have the same number of the same atoms, but they are arranged differently.

Isotopes are two atoms that have the same atomic number, but different atomic weights (caused by different numbers of neutrons in their nuclei).

Solutions are two compounds that are physically combined together, but not chemically combined.

Compounds are two atoms that are combined together in a fixed ratio, like two hydrogen atoms combined with one atom of oxygen to form water.

 

Problem 5

Which term refers to how strongly an atom of an element attracts electrons in a chemical bond with an atom of a different element?

(1) entropy

(2) electronegativity

(3) activation energy

(4) first ionization energy

 

¿Cuál término se refiere a que tan fuerte un átomo de un elemento atrae electrones en un enlace químico con los átomos de un elemento distinto?

(1) entropía

(2) electronegatividad

(3) energía de activación

(4) primera energía de ionización

 

Answer and discussion:

Entropy is a measure of the randomness of atoms.

Electronegativity is a measure of how strongly an element attracts electrons in a chemical bond with an atom of a different element.

Activation energy is the minimum amount of energy that must be added to a system in order for two compounds to react with each other.

First ionization energy is the amount of energy required to remove the first electron from an atom in forming an ionic bond.

Choice (2) is the correct choice.

 

 

 

Math Problems for 8/1/16

Problem 1

An expression of the fifth degree is written with a leading coefficient of seven and a constant of six. Which expression is correctly written for these conditions?

(1) 6x5 + x4 +7

(2) 7x6 – 6x4 +5

(3) 6x7x5 + 5

(4) 7x5 + 2x2 + 6

Una expresión de quinto grado se escribe con un coeficiente principal de siete y una constante de seis. ¿Qué expresión está escrita correctamente para estas condiciones?

(1) 6x5 + x4 + 7

(2) 7x6 – 6x4 + 5

(3) 6x7 – x5 + 5

(4) 7x5 + 2x2 + 6

Answer and discussion:

To answer this question correctly, we need to remember the definition of the terms used.

Fifth degree equations have a 5 as the superscript for a term. Coefficients are the numbers located in front of variables, whereas a constant is a number not followed by a variable.

In choice (1), these are 5, 6, and 7. In choice (2), 6, 7, 5. In choice (3), 7, 6, 5. And in choice (4), 5, 7, 6. Obviously choice (4) is the correct choice.

Problem 2

If A = 3x2 + 5x – 6 and B = -2x2 – 6x + 7, then A – B equals

(1) -5x2 – 11x + 13

(2) 5x2 + 11x – 13

(3) -5x2 – x + 1

(4) 5x2 – x + 1

 

Si A = 3x2 + 5x – 6 y B=-2x2 – 6x + 7, entonces A – B es igual a

(1) -5x2 – 11x + 13

(2) 5x2 + 11x – 13

(3) -5x2 – x + 1

(4) 5x2 – x + 1

 Answer and discussion:

What we need to do is: (3x2 + 5x – 6) – (-2x2 – 6x + 7).

First we remove the parentheses, remembering that if a parenthesis is preceded by a negative (subtract) sign, the signs of each term inside is reversed, i.e., + becomes -, and – becomes +.

(3x2 + 5x – 6) – (-2x2 – 6x + 7) = 3x2 + 5x – 6 + 2x2 + 6x – 7

We now combine like terms and arrive at our answer.

3x2 + 5x – 6 + 2x2 + 6x – 7 = 5x2 + 11x – 13, choice (2).

 

Problem 3

Which value of x satisfies the equation 7/3 * (x + 9/28) = 20?

(1) 8.25

(2) 8.89

(3) 19.25

(4) 44.92

 

¿Qué valor de x satisface la ecuación 7/3 * (x + 9/28) = 20?

(1) 8.25

(2) 8.89

(3) 19.25

(4) 44.92

Answer and discussion:

There are two ways of arriving at the correct answer for this question: (1) substitute the values in each of the four choices into the equation for x; and (2) solve the equation for x and find this value in the given choices. Let’s do it both ways.

Solution path (1):

Choice (1): 7/3 * (8.25 + 9/28) = 2.33 * (8.25 + .32) = 19.97. Pretty close to 20.

Choice (2): 7/3 * (8.89 + 8/28) = 2.33 * (8.89 + .32) = 21.46. Too high.

Choice (3): 7/3 * (19.25 + 8/28) = 2.33 * (19.25 + .32) = 45.60. Even higher.

Choice (4): 7/3 * (44.92 + 8/28) = 2.33 * (44.92 + .32) = 105.41. Sky high.

Choice (1) looks like the best answer.

 

Solution path (2):

7/3 * (x + 9/28) = 20. First we multiply both sides of the equation by 3/7 so as to get rid of the 7/3. 3/7 *[7/3 * (x + 9/28) = 20]. (x + 9/28) = 60/7.

Next we remove the parentheses, getting x + 9/28 = 20/7.

Now we subtract 9/28 from both sides, getting x + 9/28 – 9/28 = 60/7 – 9/28. x = 60/7 – 9/28.

Converting the two fractions to decimals, we get x = 8.57 – .32 = 8.25. Choice (1).

 

Problem 4

Which equation has the same solution as x2 – 6x – 12 = 0?

(1) (x + 3)2 = 21

(2) (x – 3)2 = 21

(3) (x + 3)2 = 3

(4) (x – 3)2 = 3

 

¿Qué ecuación tiene la misma solución que x2 – 6x – 12 = 0?

(1) (x + 3)2 = 21

(2) (x – 3)2 = 21

(3) (x + 3)2 = 3

(4) (x – 3)2 = 3

Answer and discussion:

In answering this question, we would simply rearrange each of the choices so that there are no binomials to be multiplied and each is equal to zero.

In choice (1), (x + 3)2 = 21. x2 + 3x + 3x + 9 = 21. x2 + 6x + 9 – 21 = 21 – 21. x2 + 9x -12 = 0. Nope.

In choice (2), (x – 3)2 = 21. x2 – 3x – 3x + 9 = 21. x2 – 6x + 9 – 21 = 21 – 21. x2 – 6x – 12 = 0. Yup.

In choice (3), (x + 3)2 = 3. x2 + 3x + 3x + 9 = 3. x2 + 6x + 9 – 3 = 3 – 3. x2 + 6x +6 = 0. Nope.

In choice (4), (x – 3)2 = 3. x2 – 3x – 3x + 9 = 3. x2 – 6x + 9 – 3 = 3 – 3. x2 – 6x + 6 = 0. Nope.

 

Problem 5

Given: L = √2

M = 3√3

N = √16

P = √9

Which expression results in a rational number?

(1) L + M

(2) M + N

(3) N + P

(4) P + L

Dado: L = √2   6/14

M = 3√3

N = √16

P = √9

¿Qué expresión da como resultado un número racional?

(1) L + M

(2) M + N

(3) N + P

(4) P + L

 

Answer and discussion:

Remember the definition of a rational number: a number that can be express as a fraction. (Note: a whole number, e.g., 3, can have a denominator of 1. Hence, 3 = 3/1.)

Now let’s look at the values of the four variables, L, M, N, and P.

L = √2. √2 is an irrational number. Therefore, any mathematical operation involving it would result in an irrational number.

 M = 3√3. Since √3 is an irrational number, all operations involving it would also be an irrational number. Hence 3√3 is irrational.

 N = √16. √16 is a rational number, namely ± 4. All operations involving it would be rational.

 P = √9. √9 is a rational number, namely ± 3. All operations involving it would be rational.

Therefore, choice (3) would be the correct choice, since it only involves rational numbers.

 

WELCOME !!!!

Beginning August 1, 2016 …

Every Monday I will post an article on how to solve a particular type of Math problem. Every Wednesday I will do the same for a Chemistry problem. Friday is Physics problem’s day.

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