# ACT Math: Five Solutions (2/27/15)

Hopefully you were able to find some time and a quiet place this weekend to try our Five Problems for Your Weekend (2/27/15). Below you’ll find the fully worked-out solutions to those problems.

9. D
In order to find f(-3), you want to plug -3 into the function in place of x. Then evaluate the expression, paying close attention to order of operations.

f(x) = 5x² + 3x – 7
f(-3) = 5(-3)² + 3(-3) – 7
f(-3) = 5(9) + 3(-3) – 7
f(-3) = 45 – 9 – 7
f(-3) = 29

20.  G
To solve this system of equations, simply add the two equations together.  This will eliminate the variable y and allow you to solve for x.

x + y = 37
xy = 13
____________
2x = 50
x = 25

Now choose one of the original equations, replace x with 25, and solve for y.

x + y = 37
25 + y = 37
y = 12

35. C
The longest side of a triangle can’t be so long that the other two sides wouldn’t be able to connect to form the triangle.  Picture sides of, for example, 2, 3, and 50. If you attach the sides of length 2 and 3 to the end points of the side with length 50, they don’t even come close to connecting. So how do we find out what the third side can be if we know two of the side lengths? Let’s call the two known sides a and b. Then the range of possible values for the third side (let’s call it c) is given by:

a – bcab

So for our problem:

22 – 8 < c < 22 + 8
14 < c < 30

The only one of the answer choices that falls in this range is 16.

48.  H
This problem tests your knowledge of function transformations, which are summarized in the table below:

transformation change in the graph of f(x)
f(x + a) moves the graph "a" units left
f(x – a) moves the graph "a" units right
f(x) + a moves the graph "a" units up
f(x) – a moves the graph "a" units down
a·f(x) stretches the graph vertically by a factor of "a"
-f(x) reflects the graph across the x-axis

Since g(x) is the function f(x) shifted 4 units right and 5 units up, we can conclude that it is true that g(x) = f(x – 4) + 5 or g(x) = (x – 4)² + 5.

51.  C
This question tests your knowledge of exponent rules.  In particular, you need to recall the Product Rule for Exponents and the rule for zero as an exponent.

name of rule rule
Product Rule (x^a)·(x^b) = x^(a + b)
Zero Exponent Rule (anything)^0 = 1

(Note that zero raised to the zero power is undefined. That’s why the problem said “for all nonzero values of x.”)

Let’s now apply the rules to our problem. $x^{k^{2}}\cdot x^{7k}\cdot x^{12}=1$ $x^{k^{2}+7k+12}=x^{0}$ $k^{2}+7k+12=0$ $(k + 4)(k + 3) = 0$ $k = -3; k = -4$

If you have questions about these problems or anything else to do with the ACT, leave a comment below or send me an email at info@cardinalec.com.