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4 years ago

Monique designed a square patio. She made the sides of the patio 15 feet. What is the area of the patio?

Mathematics

1 answer:

mixas84 [53]4 years ago

8 0

Answer:

225 square feet

Step-by-step explanation:

15feet times 15 feet.... thats how you get the area of the square

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What is 3(x-1)<12 or x+7>10

Otrada [13]

Simplifying

3[x + -1] = 12

Reorder the terms:

3[-1 + x] = 12

[-1 * 3 + x * 3] = 12

[-3 + 3x] = 12

Solving

-3 + 3x = 12

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '3' to each side of the equation.

-3 + 3 + 3x = 12 + 3

Combine like terms: -3 + 3 = 0

0 + 3x = 12 + 3

3x = 12 + 3

Combine like terms: 12 + 3 = 15

3x = 15

Divide each side by '3'.

x = 5

Simplifying

x = 5

6 0

3 years ago

IF A= -35 , B = 10 , C= -5 verify that:- a x (b+c) = a x b + a x c Plz tell

Luda [366]

Answer:

see below

Step-by-step explanation:

a x (b+c) = a x b + a x c

Let A= -35 , B = 10 , C= -5

-35 * ( 10 -5) = -35 *10 + -35 * -5

-35 *(5) = -350 + 175

-175 = -175

6 0

3 years ago

A rocket is shot straight up into the air with an initial velocity of 500 ft per second and from a height of 20 feet above the g

iragen [17]

The height of the rocket above the ground after $t$ seconds is given by the equation : $H= -16t^2+Vt+h$ , where $V$ is the initial velocity and $h$ is the initial height.

Given that, $V= 500 ft/second$ and $h= 20 ft$

So, the equation will become: $H= -16t^2 +500t+20$

A) For finding the height of the rocket 3 seconds after the launch, we will plug $t=3$ into the above equation. So....

$H= -16(3)^2+500(3)+20\\ \\ H= -16(9)+1500+20\\ \\ H= -144+1500+20=1376$

So, the height of the rocket 3 seconds after the launch is 1376 feet.

B) When the rocket at a height of 400 feet, then we will plug $H= 400$

$400=-16t^2+500t+20\\ \\ 16t^2-500t-20+400=0\\ \\ 16t^2-500t+380=0\\ \\ 4(4t^2-125t+95)=0\\ \\ 4t^2-125t+95=0$

Using quadratic formula, we will get......

$t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(95)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{14105}}{8}\\ \\ t= 30.4705... \\ and \\ t= 0.7794...$

So, after 0.7794...seconds and 30.4705...seconds the rocket is at a height of 400 feet above the ground.

C) The time duration that the rocket remains in the air means we need to find the time taken by the rocket to reach the ground. When it reaches the ground, then $H=0$ . So.....

$0=-16t^2+500t+20\\ \\ -4(4t^2-125t-5)=0\\ \\ 4t^2-125t-5=0$

Using quadratic formula, we will get.....

$t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(-5)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{15705}}{8}\\ \\ t=31.2899...\\ and\\ t= -0.0399...$

(Negative value is ignored as time can't be in negative)

So, the rocket will remain in the air for 31.2899... seconds.

5 0

4 years ago

Which set of numbers represents the domain of the set of order pairs (2,10), (-3,-10), (4,-1),(-5,2)

shutvik [7]

-3.4.-5 must be the answer

6 0

3 years ago

On a coordinate grid, point T is at (2, −4) and point S is at (2, 6). The distance (in units) between points T and S is ______.

-BARSIC- [3]

The answer is 10
This is because both points have the same x value, which is 2, which means the distance between them is a straight line. That means that we can simply find the distance between the y values, and -4 to 0 is 4 units, and 0 to 6 is 6 units, so 4 + 6 = 10.

I hope this helps!

6 0

3 years ago

Read 2 more answers