Formula for Perimeter of Rectangle:
P = 2(L + W)
Plug in 160:
160 = 2(L + W)
L = 4W
<span>So we can plug in '4W' for 'L' in the first equation.</span>
<span>160 = 2(L + W)
160 = 2(4W + W)
Combine like terms:
160 = 2(5W)
160 = 10W
Divide 10 to both sides:
W = 16
Now we can plug this back into any of the two equations to find the length.
L = 4W
L = 4(16)
L = 64
So the width is 16, and the length is 64.</span>
The anwser is (B) hope this helped
Answer:
Below.
Step-by-step explanation:
f) (a + b)^3 - 4(a + b)^2
The (a+ b)^2 can be taken out to give:
= (a + b)^2(a + b - 4)
= (a + b)(a + b)(a + b - 4).
g) 3x(x - y) - 6(-x + y)
= 3x( x - y) + 6(x - y)
= (3x + 6)(x - y)
= 3(x + 2)(x - y).
h) (6a - 5b)(c - d) + (3a + 4b)(d - c)
= (6a - 5b)(c - d) + (-3a - 4b)(c - d)
= -(c - d)(6a - 5b)(3a + 4b).
i) -3d(-9a - 2b) + 2c (9a + 2b)
= 3d(9a + 2b) + 2c (9a + 2b)
= 3d(9a + 2b) + 2c (9a + 2b).
= (3d + 2c)(9a + 2b).
j) a^2b^3(2a + 1) - 6ab^2(-1 - 2a)
= a^2b^3(2a + 1) + 6ab^2(2a + 1)
= (2a + 1)( a^2b^3 + 6ab^2)
The GCF of a^2b^3 and 6ab^2 is ab^2, so we have:
(2a + 1)ab^2(ab + 6)
= ab^2(ab + 6)(2a + 1).
Answer:
18
Step-by-step explanation:
-3-6= -8 but since its absolute value it'll be 8 so 8 plus 10 = 18
In general, you cannot prove that because it is not true.
2 is rational. √2 is not.
