Let
x-------> the width of the rectangular area
y------> the length of the rectangular area

we know that
y=x+15------> equation 1
perimeter of a rectangle=2*[x+y]
2x+2y <= 150-------> equation 2

substitute 1 in 2
2x+2*[x+15] <=150--------> 2x+2x+30 <=150----> 4x <=150-30
4x <= 120---------> x <= 30
the width of the rectangular area is at most 30 ft
y=x+15
for x=30
y=30+15------> y=45
the length of the rectangular area is at most 45 ft

see the attached figure
the solution is<span> the shaded area</span>

4 0

2 years ago

Find the inverse of 2∛(x+2)

Artemon [7]

$y=2\sqrt[3]{x+2}\ \ \ |divide\ both\ sides\ by\ 2\\\\\sqrt[3]{x+2}=\dfrac{y}{2}\to x+2=\left(\frac{y}{2}\right)^3\ \ \ |subtract\ 2\ from\ both\ sides\\\\x=\dfrac{y^3}{2^3}-2\\\\x=\dfrac{1}{8}y^3-2\\\\\boxed{f(x)=2\sqrt[3]{x+2}\to f^{-1}(x)=\frac{1}{8}y^3-2}$

5 0

3 years ago

3y +(-2) +48=180 solve for y

GalinKa [24]

Answer:

The answer to the problem is

y = 134/3

Step-by-step explanation

7 0

3 years ago