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NISA [10]
3 years ago
14

HELP 30 POINTS!!!!!HELP 30 POINTS!!!!!HELP 30 POINTS!!!!!HELP 30 POINTS!!!!!HELP 30 POINTS!!!!!

Mathematics
2 answers:
rewona [7]3 years ago
4 0

Answer:

-10 if you meant 1/4(8-4p)

-20 if you meant 1/4{(8-4p)+2p}

Step-by-step explanation:

hope this helps have a great day.

cupoosta [38]3 years ago
3 0

Answer:

-10 if you meant 1/4(8-4p)

-20 if you meant 1/4{(8-4p)+2p}

Step-by-step explanation:

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PPLEASEE HELP<br> FIND THE VOLUME
Doss [256]

Answer:

volume = 7 * 4 * 24 = 672 m 3

Step-by-step explanation:

a^2 = 25^2 - 7^2 = 576

-> a = 24

volume = 7 * 4 * 24 = 672 m 3

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3 years ago
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25m/s to miles/h <br>Can anyone solve it?<br>With explanation if you can
Dimas [21]

25 meters per second x 60 seconds = 1500 meters per minute

1500 meters per minute x 60 minutes per hour = 90,000 meters per hour

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90000/1609.34 = 55.92 miles per hour

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3 years ago
For g (2) = 3x – 2 find the value of x<br> -<br> for which g(x) = 10
hram777 [196]

Answer:

For g (2) = 3x -2 find the value of x

The value of x is 2

for which g(x) = 10?

10 = 3x -2

10 +2 = 3x

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x = 12/3

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3 0
3 years ago
You score an 88 on the first of two exams. Write and solve an equation to find the score x that you need on the second exam to h
harkovskaia [24]

Answer:

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Step-by-step explanation:

6 0
3 years ago
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Please help, the question below
Licemer1 [7]

Answer:

\sf R=\left(3, -\dfrac{5}{4}\right)

Step-by-step explanation:

Given:

  • P = (1, 5)
  • Q = (-2, 3)
  • R = (a, b)
  • R lies on line x = 3
  • PR = QR

If point R lies on the line x = 3, then the x-value of point R is 3.

⇒ a = 3

<u />

<u>Distance between two points</u>

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

(where (x₁, y₁) and (x₂, y₂) are the two points)

Use the <u>distance formula</u> to derive equations for PR and QR.

Let (x₁, y₁) = P = (1, 5)

Let (x₂, y₂) = R = (3, b)

\begin{aligned} \sf PR & =\sf \sqrt{(3-1)^2+(b-5)^2}\\ & = \sf \sqrt{4+(b-5)^2} \end{aligned}

Let (x₁, y₁) = Q = (-2, 3)

Let (x₂, y₂) = R = (3, b)

\begin{aligned} \sf QR & =\sf \sqrt{(3-(-2))^2+(b-3)^2}\\ & = \sf \sqrt{25+(b-3)^2} \end{aligned}

As PR = QR, equate the derived equations and solve for b:

\begin{aligned} \sf PR & = \sf QR \\\sf \sqrt{4+(b-5)^2} & = \sf \sqrt{25+(b-3)^2}\\\sf 4+(b-5)^2 & = \sf 25+(b-3)^2\\\sf 4+b^2-10b+25 & = \sf 25 + b^2-6b+9\\\sf b^2-10b+29 & = \sf b^2 -6b +34\\\sf -10b+29 & = \sf -6b + 34\\\sf -4b & = \sf 5\\\sf b & = -\dfrac{5}{4}\end{aligned}

Substitute the found values of a and b to find the coordinates of R:

\sf R=(a,b)=\left(3, -\dfrac{5}{4}\right)

Learn more about the distance formula here:

brainly.com/question/28144723

8 0
2 years ago
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