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3 years ago

The area of a square bedroom is 144 square feet. what is the length of one wall?

1 answer:

6 0

The answer is: "12 feet" .
________________________________
Note: In a square, the length of EACH of the four sides of the square is the same.

Area = Length * width.

For a square, length = width.

So for a square, Area = length * width = (length of a side)² = s² ,

Given: A = s² = 144 ft² ;

Solve for the positive value of "s" .
________________________________
  → s² = 144 ft² ; Take the "square root" of each side ;

  → √(s²) = √(144 ft²) ;

  → s = 12 ft.
_______________________________
The answer is: 12 ft.
_______________________________

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Write an exponential function to model the following situation

Dominik [7]

Answer:

$y = 140000(1.04)^{x}$ ; 262 000

Step-by-step explanation:

The general formula for an exponential function is

$y = a(b)^{x}$

a = 140 000; b = 1.04; x = 16

$y = 140000(1.04)^{x}$

y = 140 000(1.04)¹⁶

y = 140 000 × 1.873

y = 262 000

The graph below shows the population growth over time.

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3 years ago

Find the value of x that makes each sentence true.

ki77a [65]

Answer:

51/25 and 27/4

Step-by-step explanation:

distributing and dividing both sides by coefficient of x

8 0

3 years ago

Read 2 more answers

A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$

= $\int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt$

= By using the Scientific calculator notation;

distance = 106.109 m

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3 years ago

Ken’s predicted temperature for April is 7ºC. His prediction for May is 2ºC higher than April. He used the number line to solve

Gekata [30.6K]

Answer:

He should have moved to the right.

Step-by-step explanation:

Ken moved 7 to the right which is correct because he needs to get 7º higher, but when he moved 2º to the left that would be a temperature decrease which is is incorrect.

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3 years ago

Read 2 more answers

I’m confused on this one

ad-work [718]

The question is "What is the scale of the model to the actual statue"

That means you must put 2/15.

2 being the model

15 being the actual statue

Take that fraction and equal is to 1/x.

Multiply 15 to 1 and 2 to x.

In the end, to solve, you divide 15 by 2x.

The answer is X=7.5

1 : 7.5

7 0

3 years ago