Can someone please Help me?

Question

1)sqrt y+ sqrt y+7=7

2)sqrt x+11= sqrt x +1

3)Find the dimensions of a rectangle a with the greatest area whose perimeter is 30ft

4)I y varies directly as x and inversely as z^2, and y =3 when x=2 and z =4, what is the value of x when y=9 and z=4?

Answer 1

1) √[y] + √[y] + 7 = 7

2√[y] = 7 - 7

2√[y] = 0

2√[y] / 2 = 0 / 2

√[y] = 0

(√[y])² = 0²

y = 0

Or did you mean?: √[y] + √[y+7] = 7

√[y+7] = 7 - √[y]

(√[y+7])² = (7 - √[y])²

y + 7 = (7 - √[y])(7 - √[y])

y + 7 = 49 -14√[y] + y

-42 = -14√[y]

(-42)/(-14) = √[y]

3 = √[y]

3² = (√[y])²

y = 9


2) √[x+11] = √[x + 1]

x + 11 = x + 1

no solution

Or did you mean?:

√[x] + 11 = √[x] + 1

no solution

Or did you mean?:

√[x+11] = √[x] + 1

x + 11 = (√[x] + 1)(√[x] + 1)

x + 11 = x + 2√[x] + 1

10 = 2√[x]

√[x] = 5

x = 25

3) Perimeter: P = 2L + 2W = 30 → L + W = 15

Maximize: Area: A = LW
A = LW
W = 15 - L

A = L(15 - L)

A = 15L - L²

dA/dt = 15 - 2L = 0 (to find critical points)

2L = 15

L = 7.5

A is a max at L = 7.5, and since W = 15 - L, then W = 15 - (7.5) = 7.5

So to maximize area,

L = 7.5ft and W = 7.5 ft


4) y = kx / z²

3 = 2k / 16 → 3 = k / 8, k = 24

y = 24x / z²

9 = 24x / 16

9 (2/3) = x

x = 6

Answer 2

Please learn to bracket appropriately.

1) I assume you mean √y + √(y+7) = 7, since other wise you just get √y = 0 and hence y = 0.
So we get √(y+7) = 7 - √y
<=> y + 7 = 49 - 14√y + y (squaring both sides)
<=> 14 √y = 42
<=> √y = 3
<=> y = 9.

2) Similarly, I assume this should be √(x+11) = √x + 1. The other possible combinations don't make much sense. Again, square both sides:
x + 11 = x + 2√x + 1
<=> 2√x = 10
<=> x = 25.

3) If length is l and width is w, we have 2l + 2w = 30 => l = 15 - w. The area ia A = lw = (15-w)w = 15w - w^2. Since this is a quadratic we know it will have maximum value at w = -15 / -2 = 15/2. So l = 15 - 15/2 = 15/2 also, i.e. it's a square (as we'd have expected), with sdes of length (15/2) = 7.5 ft.

4) We can shortcut this somewhat by noting that z has not changed. So y varies directly as x; y has tripled, so x has also tripled. So x = 6.
More generally, we would write y = kx/z^2; when x=2 and z=4, y=3, so we get 3 = 2k/16 => k = 24.
So when y = 9 and z = 4 we have 9 = 24x / 16 = 3x/2 => x = 2(9)/3 = 6.