Ask question

Ask question

All categories

2 years ago

14

Can someone please help me find the side length area, and please draw the shape on how it would look if it were finished

1 answer:

Mumz [18]2 years ago

6 0

The side length is 5 units, and the area is 25 square units

<h3>What are areas?</h3>

The area of a shape is the amount of space it occupies

<h3>What is the area of a square</h3>

The area of a square is the product of its dimensions

From the graph, the coordinates of the square can be assumed to be:

$A = (0,0)$

$B = (4,3)$

The side length of the square is then calculated using the following distance formula

$d = \sqrt{(x_2- x_1)^2 + (y_2 -y_1)^2}$

So, we have:

$AB = \sqrt{(0 - 4)^2 + (0-3)^2}$

$AB = \sqrt{25}$

$AB = 5$

The area of the square is the calculated as:

$Area = AB^2$

This gives

$Area = 5^2$

$Area = 25$

Hence, the side length is 5 units, and the area is 25 square units

See attachment for the diagram of the shape

Read more about areas at:

brainly.com/question/24487155

You might be interested in

What is the answer to 7+y<12

LiRa [457]

Step-by-step explanation:

7 + y < 12 it can also be written as

7 + y = 12

y = 12 - 7

y = 5

or

y < 5

Hope it will help :)

3 0

3 years ago

It costs 9.95 for 1 ticket to the movies. If 3 people go, how much would the total price of 3 tickets be?

SIZIF [17.4K]

It would be $9.95 x 3

4 0

3 years ago

Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane

liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

Answer:

$Rate = 0.935042^\circ /cm$

Step-by-step explanation:

Given

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$T(x,y) =x\sin2y$

$r = 1m$

$v = 2m/s$

Express the given point P as a unit tangent vector:

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$u = \frac{\sqrt 3}{2}i - \frac{1}{2}j$

Next, find the gradient of P and T using: $\triangle T = \nabla T * u$

Where

$\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})} = (sin \sqrt 3)i + (cos \sqrt 3)j$

So: the gradient becomes:

$\triangle T = \nabla T * u$

$\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] * [\frac{\sqrt 3}{2}i - \frac{1}{2}j]$

By vector multiplication, we have:

$\triangle T = (sin \sqrt 3)* \frac{\sqrt 3}{2} - (cos \sqrt 3) \frac{1}{2}$

$\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)$

$\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5$

$\triangle T = 0.935042$

Hence, the rate is:

$Rate = \triangle T = 0.935042^\circ /cm$

3 0

2 years ago

This is the graph of y=x^2+2x-2 what is the range of this function

Sedaia [141]

The range is [-3, ∞)

3 0

3 years ago

During their lastest visit to the eye doctor, Ebony and Chelsea got their eyes checked.

frutty [35]

1. Ebony's right eye....opposite of -1.50 = 1.50
2. the rational numbers are : -1.50, 1.50, -1.00, and 0.25....these are all rational because they can be turned into fractions.
3. least to greatest : -1.50, -1.00, 0.25, 1.50
4. not sure how to answer this...negatives are always less then positives....however, with negatives,unlike positives, the larger the number, the smaller it is.

7 0

3 years ago