3 years ago

Need help!! Need to solve this sheet that is attached Please help! 100 points

Mathematics

1 answer:

Mama L [17]3 years ago

5 0

Answer:

I can't see your picture

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The perimeter of a rectangle is 400 yards what are the dimensions of the rectangle if the length is 40 yards more than the with

zloy xaker [14]

P = 2(L + W)
P = 400
L = W + 40

400 = 2(W + 40 + W)
400 = 2(2W + 40)
400 = 4W + 80
400 - 80 = 4W
320 = 4W
320/4 = W
80 = W.....so the width (W) = 80 yards

L = W + 40
L = 80 + 40
L = 120...and the length (L) = 120 yards

7 0

3 years ago

Please anyone solve this for me..No spam.

Tamiku [17]

Answer:

3096

Step-by-step explanation:

Given:

$3x-\frac{1}{4y}=24$

Rewrite it as:

$3x=\frac{1}{4y}+24$
$x=\frac{1}{12y}+8$

Simplify the expression:

$8x(6x - \frac{1}{y} ) + \frac{1}{12y^2} (4 - 3y + 36xy^2) =$
$8(\frac{1}{12y}+8)(6(\frac{1}{12y}+8) - \frac{1}{y} ) + \frac{1}{3y^2} - \frac{1}{4y} + 3x =$
$8(\frac{1}{12y}+8)(48- \frac{1}{2y} ) + \frac{1}{3y^2} - \frac{1}{4y} + 24 + \frac{1}{4y} =$
$8(\frac{4}{y}+8*48- \frac{1}{24y^2} -\frac{4}{y} ) + \frac{1}{3y^2} + 24 =$
$8(384-\frac{1}{24y^2}) + \frac{1}{3y^2} + 24 =$
$8*384-\frac{1}{3y^2} + \frac{1}{3y^2} + 24 =$
$3072+24=$
$3096$