# ACT Math: Five Solutions (2/6/15)

Hopefully you were able to find some time and a quiet place this weekend to try our Five Problems for Your Weekend (2/6/15). Below you’ll find the fully worked-out solutions to those problems.

17.  E
Begin by finding the slope of the line using the Slope Formula:

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{9-3}{1-(-2)} = \frac{6}{3} = 2$

Once you have the slope, you can plug that in to the point-slope formula:

$y-y_{1} = m(x-x_{1})$

Finally, rearrange that into y = mx + b to find the y-intercept, which is represented by b.

y – 9 = 2(x – 1)
y – 9 = 2x – 2
y – 9 + 9 = 2x – 2 + 9
y = 2x + 7

The y-intercept of the line is 7.

26.  J
You’ll recall from Algebra 1 that the equation of any non-vertical line can be written in the form y = mx + b, where m is the slope of the line and b is the y-intercept.

9x – 6y + 15 = 0
9x – 6y = -15    ← Subtract 15 from each side
-6y = -9x – 15     ← Subtract 9x from each side
y = -9/-6 x – 15/-6    ← Divide both sides by -6
y = 3/2 x + 5/2   ← Reduce the fractions

The slope of the line is 3/2.

35.  B
You can find the distance using the Distance Formula:

$d = \sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}$
$d = \sqrt{(12-7)^{2}+(1-(-3))^{2}}$
$d = \sqrt{5^{2}+ 4^{2}}$
$d = \sqrt{25 + 16}$
$d = \sqrt{41}$

If you didn’t remember the Distance Formula, you can still get the problem correct by sketching it out, drawing a right triangle and using the Pythagorean Theorem (the Distance Formula is really Pythagorean Theorem “in disguise”).

$a^{2} + b^{2} = c^{^{2}}$
$4^{2} + 5^{2} = c^{^{2}}$
$16 + 25 = c^{^{2}}$
$41 = c^{^{2}}$
$\sqrt{41} = c$

You’ll get the same answer using either the Distance Formula or by using Pythagorean Theorem. Use the method that makes the most sense to you.

42.  H
You’ll need to find the area of each of the three circles in the diagram using the formula for the area of a circle:  A = πr².

Large Circle:  r = 6 and A = π(6)² = 36π
Medium Circle:  r = 4 and A = π(4)² = 16π
Small Circle:  r = 2 and A = π(2)² = 4π

The total of the shaded regions is equivalent to:

Area of Large Circle – Area of Medium Circle + Area of Small Circle
36 π – 16π + 4π = 24π

57.  C
Begin by sketching the triangle on the xy-coordinate plane. Then use the Pythagorean Theorem to find the missing third side of the triangle.

$a^{2} + b^{2} = c^{^{2}}$
$9^{2} + b^{2} = 41^{^{2}}$
$81 + b^{2} = 1681$
$b^{2} = 1600$
$b = 40$

As you label the triangle, pay attention to the signs of the sides as well as their lengths. Because the triangle is in the third quadrant, where 180 < θ < 270, both the horizontal and vertical sides of the triangle should be labeled with negative values.  Now you can find the value of sin θ, which is the ratio of the opposite side to the hypotenuse. That value is -40/41.

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 .