HELP: Subsets Attached is the photo reference

We call x the 4 points-worth problems and y the 3 points-worth problems

You know that x+y = 32

4x+3y = 111

You know that the difference from the x and y is 32 so write:

x = 32-y

Substitute at x the value of 32-y

4(32-y)+3y = 111

128-4y +3y = 111

-4y+3y = 111-128

-y = -17

y = 17

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3 years ago

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Does anyone have the answers for all?

Shkiper50 [21]

Answer and Step-by-step explanation:

1-6

a.

What you can do is plug in 6 and start by doing the first machine first, then the second. If this doesn't work, do the second machine, then the first machine.

The order that works is second machine, then first machine.

Second Machine:

y = $(6)^{2} -6$ = 36 - 6 = 30

Plug into first machine now.

First Machine:

y = $\sqrt{(30) - 5} = \sqrt{25} = 5$

We see that the result is 5.

b. If we were to do the order of second machine, then first machine, we will never get a negative result (or -5 for that matter), due to the fact that you can't algebraically square root a negative number.

However, if we were to do the order of first machine, second machine, we can get a negative number. To get -5, we plug in 6 for x in the first machine, getting an answer of 1. We plug this into the second machine, getting 1^2 - 6, which results in -5.

1-7. The absolute value of a value is the positive version of that value.

a. |54| = 54

b. $-|-7\frac{3}{5}|$ = $-7\frac{3}{5}$

c. |3| - |-1| = 3 - 1 = 2

d. |2.2 - 5.13| = |-2.93| = 2.93

1-8.

a.

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(5 boxes on each side)

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(6 boxes on each side)

b. The amount of squares on each side of the figure increases by 1.

c. There would be 1 square, because we know that the pattern is there is 2 squares added to the next figure, 1 at each end. If we go back a figure, we take away 2 squares.

1-9.

a. -42 - 17 = -59

b. 8 - (-9) = 8 + 9 = 17

c. 8(-9) = -72

d. -42 ÷ (-7) = 6

e. -2(-3)(-4) = 6(-4) = 24

f. -18 - 7 = -25

g. $(-5)^{2}$ = 25