Hopefully you were able to find some time and a quiet place this weekend to try our Five Problems for Your Weekend (3/20/15). Below you’ll find the fully worked-out solutions to those problems.

15. B

Begin by re-arranging 4*x* – 8*y* = 24 into slope-intercept form.

4*x* – 8*y* = 24

-8*y* = -4*x* + 24

*y* = 1/2 *x* – 3

As you may recall from algebra, perpendicular lines have slopes that are opposite reciprocals. Because the slope of the original line is 1/2, we are looking for a line with a slope of -2. That makes the correct answer Choice B.

32. K

This is a tricky little problem — it depends upon your understanding of factoring and exponents rules. Start by taking out a Greatest Common Factor (GCF) and then substitute 5 for *ab* in a couple of places.

*a*²*b*²* + a*³*b*³

*a*²*b²*(1* + ab*)

(*ab*)²(1 + *ab*)

(5)²(1 + 5)

25(6)

150

One of the keys to getting this problem correct is to recognize that *a*²*b² *is equivalent to (*ab*)². You really need to know your exponent rules well.

41. A

You’ll need to use Pythagorean Theorem twice to find the lengths of both segment *BD* and segment *DC*.

24² + (*BD*)² = 26²

576 + (*BD*)² = 676

(*BD*)² = 100

*BD = *10

10² + (*DC*)² = (√149)²

100 + (*DC*)² = 149

(*DC*)² = 49

*DC* = 7

You can now determine that the length of segment *AC*, the base of the triangle, is 31. The area of the triangle then is ½(31)(10) = 155 square units.

44. F

This is a right triangle trigonometry (SOH-CAH-TOA) problem. The ramp is the hypotenuse of the right triangle and we are looking for the side opposite the given angle, so we can write an equation involving sine.

sin 3.8 = *x*/20

20(sin 3.8) = *x
*

*x*≈ 1.3

For the final calculation, be sure that your calculator is set to degrees and not radians.

53. C

Because the tangent of ∠ *A* is 2/3, you know the ratio of the opposite side to the adjacent side is 2 to 3. Those sides can be labeled 2*x* and 3*x*.

You can now use the Pythagorean Theorem to find the value of *x*.

(2*x*)² + (3*x*)² = (√117)²

4*x**² + *9*x*² = 117

13*x*² = 117

*x**² = *9

*x* = 3

Because segment *CB* has been represented by 2*x, *its length is 2(*x*) = 2(3) = 6.

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.

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