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Nostrana [21]
2 years ago
12

A rectangle has a length of 2x+7 and a width of 3x+8.

Mathematics
1 answer:
Phantasy [73]2 years ago
3 0

Answer:

Below.

Step-by-step explanation:

Perimeter = 2(2x + 7 + 3x + 8)

= 2(5x + 15)

= 10x + 30.

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Find the base and height of a rectangle with vertices (-4,7), (-2,7), (-2,1), and (-4,1).
Salsk061 [2.6K]

Answer:

The height is 6 unit

The base is 2 units.

Step-by-step explanation:The given rectangle has vertices (–4, 7), (–2, 7), (–2, 1), and (–4, 1).

3 0
3 years ago
Being fun again who whants points here is a question you'll get 50 points
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Answer:

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8 0
3 years ago
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The graph below shows the amount of money left in the school’s desk fund, f, after d desks have been purchased. A graph titled D
natima [27]

Answer:

$110

Step-by-step explanation:

The computation is shown below:

Provided that

Number of desk purchased    Fund Left $

           0                                   700

           1                                    590

           2                                   480

           3                                   370

           4                                   260

           5                                   150

           6                                    40

Now the Fund decreased while purchasing is

= 5 desks - 2 desks

= 3 desks

i.e

= $480 - $150

= $330

Now the amount of money left for decreased the desk fund is

= \frac{Fund\ decreased }{number\ of\ desk\ decreased}

= \frac{\$330}{3\ desk}

= $110

3 0
3 years ago
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Inequality to x2 < 64
Anarel [89]
              NOT MY WORDS TAKEN FROM A SOURCE!

(x^2) <64  => (x^2) -64 < 64-64 => (x^2) - 64 < 0 64= 8^2    so    (x^2) - (8^2) < 0  To solve the inequality we first find the roots (values of x that make (x^2) - (8^2) = 0 ) Note that if we can express (x^2) - (y^2) as (x-y)* (x+y)  You can work backwards and verify this is true. so let's set (x^2) - (8^2)  equal to zero to find the roots: (x^2) - (8^2) = 0   => (x-8)*(x+8) = 0       if x-8 = 0 => x=8      and if x+8 = 0 => x=-8 So x= +/-8 are the roots of x^2) - (8^2)Now you need to pick any x values less than -8 (the smaller root) , one x value between -8 and +8 (the two roots), and one x value greater than 8 (the greater root) and see if the sign is positive or negative. 1) Let's pick -10 (which is smaller than -8). If x=-10, then (x^2) - (8^2) = 100-64 = 36>0  so it is positive
2) Let's pick 0 (which is greater than -8, larger than 8). If x=0, then (x^2) - (8^2) = 0-64 = -64 <0  so it is negative3) Let's pick +10 (which is greater than 10). If x=-10, then (x^2) - (8^2) = 100-64 = 36>0  so it is positive Since we are interested in (x^2) - 64 < 0, then x should be between -8 and positive 8. So  -8<x<8 Note: If you choose any number outside this range for x, and square it it will be greater than 64 and so it is not valid.

Hope this helped!

:)
7 0
3 years ago
What is the length of the hypotenuse, x, if (20,21,x) is a Pythagorean triple?
Sophie [7]

Answer:

<h2>29</h2>

Step-by-step explanation:

(20,21,x) \\Pythagoras -Theorem =\\a^2+b^2 = c^2\\\\20^2 + 21^2 = x^2\\400 + 441 = x^2\\841 = x^2\\Square -root-both-sides\\\sqrt{841} = \sqrt{x^2} \\29 = x\\\\x = 29

6 0
3 years ago
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