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olga55 [171]
3 years ago
15

Gloria forgot to account for two walkways. Using the work from Plan A,

Mathematics
1 answer:
marissa [1.9K]3 years ago
4 0

Answer:

hhs a

Step-by-step explanation:

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Point M is located at (7,1).
Charra [1.4K]

I think it is possibly point b

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3 years ago
50a3 + 10a2
Lady_Fox [76]
50a^3+10a^2\\\\50a^3=10\cdot5\cdot a^2\cdot a=10a^2\cdot5a\\\\50a^3+10a^2=10a^2\cdot5a+10a^2\cdot1=\huge\boxed{10a^2(5a+1)}\to\boxed{A}
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3 years ago
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You wanna answer it you get SOME POINTS . AND YOU GET TO MARRY THE GIRL OF YOU'R DREAMS ! ! ! IK right . So do it .
uysha [10]

Answer:

The answer to your question is 20 feet

Step-by-step explanation:

Data

A = (-2, 1)

B = (4, 1)

C = (-2, -3)

D = (4, -3)

Process

1.- Calculate the distance from C and D

dCD = \sqrt{(4 + 2)^{2} + (3 - 3)^{2}}

dCD = \sqrt{6^{2}}

dCD = 6

2.- Calculate the distance from A to C

dAC = \sqrt{(2 - 2)^{2} + (-3 - 1)^{2}}

dAC = \sqrt{4^{2}}

dAC = 4

3.- Calculate the perimeter

Perimeter = 2(dCD) + 2(dAC)

-Substitution

Perimeter = 2(6) + 2(4)

-Simplification

Perimeter = 12 + 8

-Result

Perimeter = 20 ft

3 0
3 years ago
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Which is a quadratic function
eimsori [14]

Answer:

im sorry but i think it is 4

Step-by-step explanation:

5 0
3 years ago
Special air bags are used to protect scientific equipment when a rover lands on the surface of Mars. On Earth, the function appr
PSYCHO15rus [73]

The nearest tenth of how fast a rover will hit Mars' surface after a bounce of 15 ft in height is 20.7ft/s.

<h3>What is the approximation about?</h3>

From the question:

Mars: F(x) = 2/3\sqrt{64x}

Therefore, If x = 15

Then:

f (15) =  2/3 \sqrt[8]{15}

= 16/3\sqrt{15}

= 20.7ft/s

Hence, The nearest tenth of how fast a rover will hit Mars' surface after a bounce of 15 ft in height is 20.7ft/s.

Learn more about Estimation from

brainly.com/question/24592593

#SPJ1

8 0
2 years ago
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