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creativ13 [48]
4 years ago
10

Simplify 3(3 √9)^-2 show work if possible :)

Mathematics
1 answer:
kow [346]4 years ago
6 0
3<span>(3 √9)^-2 =

3/(3</span>√9)^2 =
3/(9*9) = 
1/27
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Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
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Matching each equation of the plecewise function represented in the graph will be:

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  • 0 < x < 2 = 1
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<h3>How to illustrate the information?</h3>

It should be noted that a domain simply means the set of inputs that are accepted by the function.

In this case, the equation of the piecewise function represented is given.

The graph is attached.

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brainly.com/question/2972832

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Is this the right answer?
pychu [463]
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3 years ago
If f(x)=4x+10, find f(2)
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Answer:

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3 years ago
On the school playground, the slide is 8 feet due west of the tire swing and 15 feet due south of the monkey bars. What is the d
xeze [42]
23 you add 8 to 15, that's how u get your answer
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3 years ago
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Help please ITS OF TRIGONOMETRY<br>PROVE ​
Flauer [41]

Answer:

The equation is true.

Step-by-step explanation:

In order to solve this problem, one must envision a right triangle. A diagram used to represent the imagined right triangle is included at the bottom of this explanation.  Please note that each side is named with respect to the angle is it across from.

Right angle trigonometry is composed of a sequence of ratios that relate the sides and angles of a right triangle. These ratios are as follows,

sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}

One is given the following equation,

\frac{sin(A)+sin(B)}{cos(A) +cos(B)}+\frac{cos(A)-cos(B)}{sin(A)-sin(B)}=0

As per the attached reference image, one can state the following, using the right angle trigonometric ratios,

sin(A)=\frac{a}{c}\\\\sin(B)=\frac{b}{c}\\\\cos(A)=\frac{b}{c}\\\\cos(B)=\frac{a}{c}

Substitute these values into the given equation. Then simplify the equation to prove the idenity,

\frac{sin(A)+sin(B)}{cos(A) +cos(B)}+\frac{cos(A)-cos(B)}{sin(A)-sin(B)}=0

\frac{\frac{a}{c}+\frac{b}{c}}{\frac{b}{c}+\frac{a}{c}}+\frac{\frac{b}{c}-\frac{a}{c}}{\frac{a}{c}-\frac{b}{c}}=0

\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{b-a}{c}}{\frac{a-b}{c}}

Remember, any number over itself equals one, this holds true even for fractions with fractions in the numerator (value on top of the fraction bar) and denominator (value under the fraction bar).

\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{b-a}{c}}{\frac{a-b}{c}}

\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{-(a-b)}{c}}{\frac{a-b}{c}}

1+(-1)=0

1-1=0

0=0

7 0
3 years ago
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