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2 years ago

A new Neighborhood Activity Complex is being built-in

Mathematics

1 answer:

Anestetic [448]2 years ago

7 0

Step-by-step explanation:

L = 2W - 8

P = 230 yd

P = 2×(L+W)

230 = 2× (2W - 8 + W)

230 = 2× (3W-8)

230 = 6W-16

6W= 230+16

6W = 246

W = 246/6

W = 41

L = 2(41) -8

= 82-8

= 74

so, the dimensions of the playing field:

the length = 74 yd

the length = 74 ydthe wide = 41 yd

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A rectangular cracker has a length of 5 centimeters and an area f 20 square centimeters. Find its perimeter.

iren [92.7K]

Answer:

18cm

Step-by-step explanation:

We have the length of the rectangle but we need to find the width which we'll call w. Area is a, l is length.

a = l * w

20 = 5 * w

divide both sides by 5 to cancel the multiplication by 5 on the right side

4 = w

The rectangle is going to have two sides that are the length of l, and two sides that are the the length of w. We're looking for the perimiter p.

p = 2l + 2w

p = (2 * 5) + (2 * 4)

p = 18

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1 year ago

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3 years ago

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$x = - 6$

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3 years ago

Which of the capital letters G, I, J, L, or Q has at least one line of symmetry?

gladu [14]

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2 years ago

Please help with this and explain how to do it, I'll give 15 points

MrRissso [65]

Answer:

Option (4).

Step-by-step explanation:

Outer diameter of the hollow metallic ball = 10 centimeters

Outer radius of this ball = $\frac{10}{2}$ = 5 cm

Volume of the outer ball $V_{1}$ = $\frac{4}{3}\pi (r)^3$

= $\frac{4}{3}\pi (5)^3$

Inner radius of the hollow metallic ball = (5 - 1) = 4 cm

Volume of the inner hollow ball $V_{2}$ = $\frac{4}{3}\pi (4)^3$

Volume of the metal used in the metallic ball = $V_{1}-V_{2}$

= $\frac{4}{3}\pi (5)^3-\frac{4}{3}\pi (4)^{3}$

= $\frac{4}{3}\pi [(5)^3-(4)^3]$

Therefore, expression given in option (4) will be used to measure the volume of the hollow metallic ball.

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2 years ago