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cluponka [151]
3 years ago
13

How do you find the perimeter and area of this shape?

Mathematics
1 answer:
mixas84 [53]3 years ago
6 0

 

P = 12 + 6 + 8,5 = 18 + 8,5 = 26,5 mm

A = 12 × 4 / 2 = 48 / 2 = 24 mm²



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The sum of five times a number and 24 is seventy nine
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Graph the linear equation x= - 5​
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Step-by-step explanation:

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Solve for the variable: -8f > 56
kolbaska11 [484]
Hi there....

-8f > 56
Divide both sides by -8
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Solving systems elimination: I don’t understand how to solve at all. Please help if you can.
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The solution to given system of equations y = 3x - 9 and y = -2x + 16 are (x, y) = (5, 6)

The solution to given system of equations y = 5x + 22 and -6x - 4y = -10 are (x, y) = (-3, 7)

<em><u>Solution:</u></em>

Given that, we have to solve each system by substitution method

<em><u>Given system of equations are:</u></em>

y = 3x - 9 ----------- eqn 1

y = -2x + 16 -------- eqn 2

Substitution method is done by substituting eqn 2 in eqn 1

<em><u>Substitute the value of "y" from eqn 2 into eqn 1</u></em>

-2x + 16 = 3x - 9

Move the variables to one side and constants to other side

-2x - 3x = -9 - 16

Combine the like terms

-5x = -25

Cancel the negative sign on both sides of equation

5x = 25

x = \frac{25}{5} = 5

<h3>x = 5</h3>

<em><u>Substitute x = 5 in eqn 1</u></em>

y = 3(5) - 9

y = 15 - 9

<h3>y = 6</h3>

Thus solution to given system of equations are (x, y) = (5, 6)

<em><u>Given another system of equations are:</u></em>

y = 5x + 22 ----- eqn 1

-6x - 4y = -10 ------ eqn 2

Substitute eqn 1 in eqn 2

-6x - 4(5x + 22) = -10

-6x - 20x - 88 = -10

Move the variables to one side and constants to other side

-26x = -10 + 88

-26x = 78

<h3>x = -3</h3>

Substitute x = -3 in eqn 1

y = 5( - 3) + 22

y = -15 + 22

<h3>y = 7</h3>

Thus solution to given system of equations are (x, y) = (-3, 7)

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This is the power to which a number is raised, or the number of times it is multiplied by itself.
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Answer:

I'm guessing an exponent. I don't fully understand what the the question Is asking, but hope I helped.

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