**Answer:**

C

**Step-by-step explanation:**

To figure out which pairs lie on -2x+4y = 16, we can plug the x and y values into the equation to see if they work. We need all pairs of x and y values in a table to work for the answer to be correct

A)

x=-4, y=2

-2x+4y = -2 * (-4) + 4 * (2) = 8 + 8 = 16. This works. Note that two negatives multiplied together make a positive, and 4*2 = 8

x=0, y=8

-2x + 4y = -2 * (0) + 4 * (8) = 0 + 32 = 32. This does not work, as 32 is not 16. Note that anything multiplied by 0 is 0

B)

x=8, y=0

-2x+4y = -2 * (8) + 4 * (0) = -16. This is not equal to 16. Note that a negative multiplied by a positive is still a negative

x=2, y=3

-2x + 4y = -2 * (2) + 4 * (3) = -4 + 12 = 8. This is not equal to 16.

C)

x=0, y=4

-2x + 4y = -2 * (0) + 4 * (4) = 0 + 16 = 16. This works

x=4, y=6

-2x + 4y = (-2)* (4) + 4 * (6) = -8 + 24 = 16. This works. As both pairs of values in the table works, this is the correct answer. Nevertheless, we can check D to make sure.

D)

x=-2, y=5

-2x+4y = -2 * (-2) + 4 * (5) = 4 + 20 = 24. This is not 16

x=-8, y=0

-2x+4y = -2 * (-8) + 4 * (0) = 16. This works, but the other pair does not work, so D is incorrect

**Answer:**

area=999

**Step-by-step explanation:**

c = 78.77182

∠α = 20.045° = 20°2'43" = 0.34986 rad

∠β = 69.955° = 69°57'17" = 1.22094 rad

h = 25.3644

area=999

Answer: The answer is 50980

Step by Step Explanation:

First, temporarily assume that two letters with I are different, call them i 1 and i 2. Three "a" are also called as a 1, a 2 and a 3, and two h as h 1 and h 2. Then there are 11 * 10 * 9 * 8 * 7 = 55440 possible "words" (one of 11 is the first letter, 10 is the second, and so on). But because equal letters do the same "words," some "words" were counted twice or more. We have to deduct the number of "parasitic" counts although it is fairly small. The words that counted more than once are divided into many disjoint sets: 1) with two I but without repetitions of a and h; 2) with two h but without repetitions of a and I 3) with two a but without repetitions of I and h; 4) with three a but without repetitions of I and h; 5) with two I and two a's; 6) two i's and tree a's; 7) two i's and two h's 8) two h's and two a's; 9) two h's and one tree. The first category includes terms counted twice and its scale is (5 * 4) * (6 * 5 * 4) = 2400 (the first I stays at one of the 5 positions, the second at one of the 4, then 11-2i-1h-2a=6). So we have 2400/2 = 1200 to subtract. Group 2 gives -600 as well, and group 3 also. Group 4 gives * (6 * 5) = 1800 (5 * 4 * 3), and the terms are counted 6 times, -300. Groups 5, 7 , 8: 5 * 4 * 3 * 2 * 6 = 720 and counted four times, therefore -180. Group 6 and 9: 5 * 4 * 3 * 2 * 1 = 120, with 12 counts, -10. Altogether -(1200 * 3 + 300 + 180 * 3 + 10 * 2) = -4460.The answer will be 55440-4460 = 50980.

**Answer:**

a) b 2, 3

**Step-by-step explanation:**

**Answer:**

He has 4 10/12 wire left

**Step-by-step explanation:**