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iogann1982 [59]
3 years ago
9

Simple algebra need help!!!​

Mathematics
1 answer:
nadezda [96]3 years ago
5 0

Answer:

12.85 m

Step-by-step explanation:

First, you have to find the arc length of the sector. To do that, use the formula for arc length.

\frac{x}{360}  \times \pi \: d

Since the figure is a semicircle, we know the central angle is 180, so plug that in for x. We also know the diameter, which is 5. So plug that in for d. Once you get the arc length, add 5 to it. 5 is the diameter and is part of the perimeter

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Using the average daily balance method, calculate the new balance for a previous
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3 years ago
Simplify: (-x^2 - 7x + 10) - (8x^2 + 8x + 8).
JulijaS [17]

Answer:

-(9 x^2 + 15 x - 2)

Step-by-step explanation:

Simplify the following:

-(8 x^2 + 8 x + 8) - x^2 - 7 x + 10

Factor 8 out of 8 x^2 + 8 x + 8:

-x^2 - 7 x + 10 - 8 (x^2 + x + 1)

-8 (x^2 + x + 1) = -8 x^2 - 8 x - 8:

-8 x^2 - 8 x - 8 - x^2 - 7 x + 10

Grouping like terms, -x^2 - 8 x^2 - 7 x - 8 x - 8 + 10 = (-x^2 - 8 x^2) + (-7 x - 8 x) + (10 - 8):

(-x^2 - 8 x^2) + (-7 x - 8 x) + (10 - 8)

-x^2 - 8 x^2 = -9 x^2:

-9 x^2 + (-7 x - 8 x) + (10 - 8)

-7 x - 8 x = -15 x:

-9 x^2 + -15 x + (10 - 8)

10 - 8 = 2:

-9 x^2 - 15 x + 2

Factor -1 out of -9 x^2 - 15 x + 2:

Answer:  -(9 x^2 + 15 x - 2)

8 0
3 years ago
H=−4.9t2+25t
lina2011 [118]

ANSWER

5

EXPLANATION

The equation that expresses the approximate height h, in meters, of a ball t seconds after it is launched vertically upward from the ground is

h(t) =  - 4.9 {t}^{2}  + 25t

To find the time when the ball hit the ground,we equate the function to zero.

- 4.9 {t}^{2}  + 25t = 0

Factor to obtain;

t( - 4.9t + 25) = 0

Apply the zero product property to obtain,

t = 0 \: or \:  \:  - 4.9t + 25 = 0

t = 0 \:  \: or \:  \: t =  \frac{ - 25}{ - 4.9}

t=0 or t=5.1 to the nearest tenth.

Therefore the ball hits the ground after approximately 5 seconds.

4 0
3 years ago
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Analyze the diagram below and complete the instructions that follow.
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Step-by-step explanation:

where is the diagram

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