8(4x+6y-2z)
32x+48y-16z
Hope this helps!
A) 5000 m²
b) A(x) = x(200 -2x)
c) 0 < x < 100
Step-by-step explanation:
b) The remaining fence, after the two sides of length x are fenced, is 200-2x. That is the length of the side parallel to the building. The product of the lengths parallel and perpendicular to the building is the area of the playground:
A(x) = x(200 -2x)
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a) A(50) = 50(200 -2·50) = 50·100 = 5000 . . . . m²
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c) The equation makes no sense if either length (x or 200-2x) is negative, so a reasonable domain is (0, 100). For x=0 or x=100, the playground area is zero, so we're not concerned with those cases, either. Those endpoints could be included in the domain if you like.
Step-by-step explanation:
3(t+5) = 9
3t+15 = 9
3t = -6
t = -2
2(f-7) = -10
2f - 14 = -10
2f = 4
f = 2
-(c - 9) = 4
-c + 9 = 4
-c = -5
c = 5
-6(2t + 8) = -84
-12t - 48 = -84
-12t = -36
12t = 36
t = 3
-10 (s + 2) = -57
-10s - 20 = -57
-10s = -37
s = 3.7
7(3w + 8)/3 = -9
By cross multiplication,
7(3w + 8) = -27
21w + 56 = -27
21w = -83
w = -3.95
35/5 = (F - 32)/9
7 = (F - 32)/9
By cross multiplication,
63 = F - 32
63 + 32 = F
95 = F
<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>.</em>
Answer:
Step-by-step explanation:
5, -4
5-4=1
5(-4)=-20
