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## Question Explanations For

##
*Practice Test 2 (Calculator)*

## 1.

*a*= 8, and then 4

*a*= 4(8) = 32.

## 2.

## 3.

*f*(

*c*), then plug this into

*g*(

*x*):

*g*(

*f*(

*c*)) =

*g*(2

*c*) = 5(2

*c*) + 1 = 10

*c*+ 1.

If you chose (A), you may have evaluated *f*(*c*) and stopped there.

## 4.

*U*is 75% of

*T*, so

*U*= 0.75

*T*. You can plug this into the equation for

*V*to get

*V*= 0.05

*U*= 0.05(0.75

*T*) = 0.0375

*T*.

## 5.

## 6.

*f*(

*x*) by 4 is the same as multiplying every

*y*-value on the graph by 4, meaning that the slope will be four times greater. You can see this for yourself: 4

*f*(

*x*) = 8

*x*+ 8, which has a slope of 8 compared to the old slope of 2.

## 7.

*y*-intercept is greater than zero, a positive value must have been added to the right side of the equation, eliminating (A) and leaving you with the correct answer choice, (B).

## 8.

## 9.

## 10.

## 11.

## 13.

If you chose (C) you may have confused which line corresponded to which crab.

## 14.

## 15.

## 16.

*f*(

*x*) = –4 when

*x*= 1.

If you chose (B) you may have forgotten the negative sign and solved for the equation of the parabola outright.

## 17.

*y*

^{2}= 333. Simplifying gives 144 =

*y*

^{2}, so

*y*= 12 or

*y*= –12. Of these, only 12 is an option.

If you chose (C), you most likely found the value of *y*^{2}.

## 18.

*N*(

*p*) doubles each time

*p*increases by 1, you know that this data can be represented by an exponential equation with a base of 2. Furthermore, when

*p*= 0,

*N*(

*p*) = 1250, so (A) is correct.

## 19.

^{3}.

## 21.

*A*, each of its sides has length . The figure on the right is made up of some full sides, which have length , and some half sides, which have length . Adding these up, you can find that the perimeter is .

## 22.

^{5}= 32 times its original size.

## 23.

*x*= 85, so (A) is true. The mean is 83.6, so B) is not true and is therefore the correct answer. This calculation also confirms that (C) is true, and that (D) is true, because the new median would be 85.5.

## 24.

## 25.

## 26.

*j*= 15

*x*where

*x*is time in hours since 7:45, and

*j*is Jorge’s distance from start in kilometers. Similarly,

*i*= 65(

*x*– 2) for Isabella, since she leaves two hours later, at 9:45 AM. You can set

*i*=

*j*and solve for

*x*:

This means that Jorge will have been running for 2.6 hours before Isabella catches up to him, after she and Tom have been driving for 2.6 – 2 = 0.6 hours. In that time, Tom has driven (77)(0.6) = 46.2 kilometers; in 2.6 hours, Jorge has gone 39 kilometers. Therefore, the distance between them is approximately 39 + 46 = 85 kilometers.

If you chose (A), you may have found the distance Jorge travelled and stopped there.

## 27.

*x*is between 1 and 9. I is false because it is only true for –

*x*. II is true because it simply combines the inequalities given in the question. III is true because for any number between 1 and 9, it will be at most 4 units away from 5 on a number line.

## 28.

*y*+

*z*= 78 – 22 – 18 = 38 kJ. Since the New Mexico Chile outside (

*x*) absorbs as much light as the Creeping Fig in the house (

*z*) and the Lavender in the house (

*y*) combined, you know that

*x*=

*z*+

*y*= 38 kJ. Therefore, the total energy absorbed by the plants if they are all in the shed is 15 + 8 + 9 + 5 = 37 kilojoules, and the total energy absorbed if they are outside is 25 + 38 + 32 + 38 = 133 kilojoules. The difference between these is 133 – 37 = 96 kilojoules.

If you chose (B), you most likely found the total energy absorbed if all the plants were in the shed. If you chose (D), you most likely found the total energy absorbed if all the plants were outside.

## 29.

*B*is a right angle, and everything is symmetric, you know that

*AB*=

*BC*, and since line

*AC*= 8, this means

*BC*= 4. Also, since the circle has a diameter of 10, its radius is 5, so

*OC*= 5. You now have a right-angled triangle (

*OBC*) with a hypotenuse of length 5 and another side of length 4. You can use the Pythagorean Theorem to find that the remaining side,

*BO*, has length 3.

## 30.

*h*(

*t*) = 0, so 0 = –2

*t*

^{2}+ 10

*t*+ 100. Factoring gives –2(

*t*– 10)(

*t*+ 5), so t must be 10 or –5 seconds. Since time cannot be negative, it took ten seconds for the ball to reach the ground.

If you chose (A), you may have forgotten the minus sign and thought that *t* = 5.

## 31.

*x*, leaving you with 7 – (–1) = 8.

## 32.

*x*– 12 – 16 + 2

*x*= 4

*x*+ 4, and combine like terms to get 5

*x*– 28 = 4

*x*+ 4 or

*x*= 32.

## 33.

*b*= 10 and

*c*= 0.2

*b*. Then c = (0.2)(10/4)= 0.5. Three times 0.5 is 1.5.

## 34.

*x*+ 10 =

*x*

^{2}+ 10

*x*+ 25, then simplify and solve: .

This tells you that *x* = –3 or *x* = –5. Make sure to plug these solutions back into the original equation to ensure that they are indeed solutions. Finally, the product of –3 and –5 is 15.

## 35.

*x*+

*y*= –

*b*; rearranging gives

*y*+

*b*= 2

*x*. Since

*y*+

*b*= 5

*x*and

*y*+

*b*= 2

*x*, you know that 5

*x*= 2

*x*so

*x*= 0.

## 36.

*b*

^{2}= 1002. Then

*b*= 80, and since the ratio of the smaller triangle to the larger is 1:2, you know that

*x*=

*b*/2 = 40.