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MissTica
3 years ago
11

PLEASE HELP Find the area of the rectangle, to the nearest hundredth, with a base of 5 inches and a height = 2.5 inches. Type a

numerical answer in the space provided. Do not include units or spaces in your answers.
a0.
Mathematics
1 answer:
il63 [147K]3 years ago
5 0

Answer:

The area of a rectangle with a width of 5 inches, and a height of 2.5 inches, would have an area 12.5 inches!

Step-by-step explanation:

To find the area of a rectangle, simply multiply the width (W) by the length (L) to find the area (A) of the rectangle.

A = W x L

A = 5 inches x 2.5 inches

A = 12.5 inches

Hope this helps! :)

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Is 9 a possible solution for -2x + 4 ≥ -14?
konstantin123 [22]

Answer:

Yes

Step-by-step explanation:

-2(9)+4= -14  and -14 is equal to -14.

6 0
3 years ago
Evaluate -5x^2 − 8x + 1 when x = -3.
Yakvenalex [24]

Hello.

Answer: 70

New equation: −5(−3^2)−(8)(−3)+1


How to get the answer:

(−5)(−9)−(8)(−3)+1

=45−(8)(−3)+1

=45−(−24)+1

=69+1

=70


Have  a nice day

7 0
3 years ago
Sebastian wants to create a right triangle with side lengths 14 inches, 48 inches, and 5
allochka39001 [22]

Answer: Choice A

\text{Yes, because } 14^2+48^2 \text{ is equal to } 50^2

=======================================================

Explanation:

The pythagorean theorem is a^2+b^2 = c^2

The value of c is always the longest side, aka hypotenuse.

The order of 'a' and b doesn't matter.

So a = 14, b = 48, c = 50

We have a^2+b^2 = 14^2+48^2 = 196+2304 = 2500

We also have c^2 = 50^2 = 2500

Both sides result in 2500 which proves that 14^2+48^2 = 50^2 is a true statement. Therefore, we do have a right triangle.

7 0
2 years ago
Using the rotation R, can you create a function R(ABCD) that is equivalent to the reflection of ABCD across both the x-axis and
galben [10]

The reflection over the x-axis is given by:

R(x,y)\to(-x,y)

And the reflection over the y-axis is given by:

R(x,y)\to(x,-y)

Thus, a function that is equivalent to the reflection of ABCD across both axis would be:

R(x,y)\to(-x,-y)

6 0
1 year ago
Hello I need help with geometry
Andre45 [30]

:podriadarlo mas s acercarpocolo  mas y uelvo lo resn

5 0
3 years ago
Read 2 more answers
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