All categories

3 years ago

. The perimeter of a square is 8y – 12. The area of the square is 100 square units. What is the value of y?

Mathematics

2 answers:

evablogger [386]3 years ago

6 0

Answer:

y = 6.5 units

Step-by-step explanation:

Let L be the length of sides of the square.

Given the following data;

Perimeter of square = 8y - 12

Area of square = 100 square units

To find the value of y;

Area of a square = L²

Substituting into the formula, we have;

100 = L²

L = √100

L = 10 units

Mathematically, the perimeter of a square is given by the formula;

Perimeter of a square = 4L

Substituting into the formula, we have;

8y - 12 = 4*10

8y - 12 = 40

8y = 40 + 12

8y = 52

y = 52/8

y = 6.5 units

lesya [120]3 years ago

5 0

Answer:

y=6.5

Step-by-step explanation:

Perimeter is 8y-12

Side=(Perimeter)/4=(8y-12)/4=2y-3

Area=(Side)^2

100=(2y-3)^2

2y-3=10

y=6.5

You might be interested in

Help please I GIVE LOTS OF POINTS

stiks02 [169]

Question 1 C

Question 2 D

Question 3 B

5 0

3 years ago

3.28×3=9.84 why is this

-Dominant- [34]

3.28·3 = 9.84 is the same as:
3.28+3.28+3.28 = 9.84

Hope this helped! Good luck!

5 0

3 years ago

Use these functions to answer the questions.

Elenna [48]

Part A
We have $f\left(x\right)=\left(x+3\right)^2-1$ . To solve for the x-intercept, we set f(x) equal to 0. That is
$\left(x+3\right)^2-1=0$

$\left(x+3\right)^2=1$

Take the square root of both sides,
$x+3=1$

$x=-2$

The x-intercept is (-2,0).

To solve for the y-intercept, we set x=0. That is
$y=\left(0+3\right)^2-1=3^2-1=9-1=8$

The y-intercept is (0, 8)

The coordinates of the optimum point are actually the vertex which can be easily seen from the vertex form equation given above. The minimum point is (-3, -1).

Part B.
We have $g\left(x\right)=-2x^2+8x+3$ .
Factor out -2
$=-2\left(x^2-4x\right)+3$

Complete the square
$=-2\left(x^2-4x+4\right)+3-2\left(4\right)$

Simplify
$g(x)=-2\left(x-2\right)^2-5$

Part C
We have $h\left(t\right)=-16t^2+28t$ .
The maximum height is 12.25 feet after 0.875 seconds from the time of the jump. The dolphin will be back in the water after 1.75 seconds. The graph of the jump is shown in the photo.