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blsea [12.9K]
3 years ago
11

When converting measures, which is a true statement?

Mathematics
1 answer:
Keith_Richards [23]3 years ago
5 0

Answer:subtract

Step-by-step explanation:

xinmofvd

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gregori [183]

Answer:

-89 +4i sqrt (3)

Step-by-step explanation:

sqrt(-48) - 89

sqrt(-1) sqrt(48) - 89

We know that sqrt(-1) = ±i

±i sqrt(48) - 89

±i sqrt(16)sqrt(3) - 89

±4i sqrt(3) - 89

Taking the principal square root

-89 +4i sqrt (3)

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Can someone answer number 5 plz
Georgia [21]

Answer:

C) 113 m^2

Step-by-step explanation:

diameter/2 =radius = 12/2 = 6

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3 years ago
Find x for,<br> sin⁻¹ 4x + sin⁻¹ 3x = -<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20%7D%7B2%7D" id="TexFormula1" title="\
Novay_Z [31]
<h2>Explanation:</h2><h2></h2>

Let's solve this problem graphically. Here we have the following equation:

sin^{-1}(4x) + sin^{-1}(3x) = -\frac{\pi}{2}

So we can rewrite this as:

f(x)=sin^{-1}(4x) + sin^{-1}(3x) \\ \\ g(x)= -\frac{\pi}{2}

So the solution to the equation is the x-value at which the functions f and g intersect. In other words:

f(x)=g(x) \\ \\ sin^{-1}(4x) + sin^{-1}(3x) = -\frac{\pi}{2}

Using graphing calculator, we get that this value occurs at:

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3 0
3 years ago
Ann is adding water to a swimming pool at a constant rate. The table below shows the amount of water in the pool after different
N76 [4]

Answer:

(a)  As time increases, the amount of water in the pool increases.

     11 gallons per minute

(b)  65 gallons

Step-by-step explanation:

From inspection of the table, we can see that <u>as time increases, the amount of water in the pool increases</u>.

We are told that Ann adds water at a constant rate.  Therefore, this can be modeled as a linear function.  

The rate at which the water is increasing is the <em>rate of change</em> (which is also the <em>slope </em>of a linear function).

Choose 2 ordered pairs from the table:

\textsf{let}\:(x_1,y_1)=(8, 153)

\textsf{let}\:(x_2,y_2)=(12,197)

Input these into the slope formula:

\textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{197-153}{12-8}=\dfrac{44}{4}=11

Therefore, the rate at which the water in the pool is increasing is:

<u>11 gallons per minute</u>

To find the amount of water that was already in the pool when Ann started adding water, we need to create a linear equation using the found slope and one of the ordered pairs with the point-slope formula:

y-y_1=m(x-x_1)

\implies y-153=11(x-8)

\implies y-153=11x-88

\implies y=11x-88+153

\implies y=11x+65

When Ann had added no water, x = 0.  Therefore,

y=11(0)+65

y=65

So there was <u>65 gallons</u> of water in the pool before Ann starting adding water.

3 0
2 years ago
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Margaret [11]
The answer is probably like 8/11
5 0
3 years ago
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