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Katen [24]
3 years ago
7

Help graph questions. (20 points)

Mathematics
2 answers:
Setler79 [48]3 years ago
5 0
First, we need the total number of people that used transportation. 
8+20+20+32 = 80
Next, we can make a ratio that represents the number of people who used bikes out of the total number of people. 8/80. Next, we set this equal to the ratio x/100 with x being the percent of people who used a bike.
8/80 = x/100
Finally, we solve for x. 
(100*8)/80 = x
800/80 = x10 = x
x equals 10, meaning 10% of people surveyed used a bike to get to the park.

<span />
Romashka [77]3 years ago
3 0
15 percent
Its 15 percent for the reason that I th k it’s is
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Determine whether
lora16 [44]

Answer:

(a) and (b) are not equivalent

(c) is equivalent

Step-by-step explanation:

Given

\frac{25^m}{5}

Required

Determine an equivalent or nonequivalent expression

(a)\ 25^{m-1

We have:

25^{m-1

Apply law of indices

25^{m-1} = \frac{25^m}{25}

<em>This is not equivalent to </em>\frac{25^m}{5}<em></em>

(b)\ 25^{2m - 1}

We have:

25^{2m - 1}

Apply law of indices

25^{2m - 1} = \frac{25^{2m}}{25}

<em>This is not equivalent to </em>\frac{25^m}{5}<em></em>

<em></em>

<em />(c)\ 5^{2m-1}<em />

We have:

<em />5^{2m-1}<em />

Apply law of indices

<em />5^{2m-1} = \frac{5^{2m}}{5^1}<em />

<em />5^{2m-1} = \frac{5^{2m}}{5}<em />

<em>Evaluate the numerator</em>

<em />5^{2m-1} = \frac{25m}{5}<em />

<em />

<em>This is an equivalent expression</em>

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Reduce 24/33 to lowest terms
Paha777 [63]

24/33

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is the correct answer

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An artist is creating a large butterfly sculpture outside a museum. There is a circular dot on each wing made out of a metal rin
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3 years ago
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Given: are diameters m∠3 = <br> A) 39 <br> B) 78<br> C) 102
Blizzard [7]
I just did this a few seconds ago and the correct answer is 78. The check above the number is green which means its correct. 
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Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first
BaLLatris [955]

Answer:

\cos(x+y) goes with -\frac{\sqrt{6}+\sqrt{2}}{4}

\sin(x+y) goes with \frac{\sqrt{6}-\sqrt{2}}{4}

\tan(x+y) goes with \sqrt{3}-2

Step-by-step explanation:

\cos(x+y)

\cos(x)\cos(y)-\sin(x)\sin(y) by the addition identity for cosine.

We are given:

\sin(x)=\frac{\sqrt{2}}{2} which if we look at the unit circle we should see

\cos(x)=\frac{\sqrt{2}}{2}.

We are also given:

\cos(y)=\frac{-1}{2} which if we look the unit circle we should see

\sin(y)=\frac{\sqrt{3}}{2}.

Apply both of these given to:

\cos(x+y)

\cos(x)\cos(y)-\sin(x)\sin(y) by the addition identity for cosine.

\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}

\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}

\frac{-\sqrt{2}-\sqrt{6}}{4}

-\frac{\sqrt{6}+\sqrt{2}}{4}

Apply both of the givens to:

\sin(x+y)

\sin(x)\cos(y)+\sin(y)\cos(x) by addition identity for sine.

\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}

\frac{-\sqrt{2}+\sqrt{6}}{4}

\frac{\sqrt{6}-\sqrt{2}}{4}

Now I'm going to apply what 2 things we got previously to:

\tan(x+y)

\frac{\sin(x+y)}{\cos(x+y)} by quotient identity for tangent

\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}

-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}

-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}

-\frac{8-2\sqrt{12}}{4}

There is a perfect square in 12, 4.

-\frac{8-2\sqrt{4}\sqrt{3}}{4}

-\frac{8-2(2)\sqrt{3}}{4}

-\frac{8-4\sqrt{3}}{4}

Divide top and bottom by 4 to reduce fraction:

-\frac{2-\sqrt{3}}{1}

-(2-\sqrt{3})

Distribute:

\sqrt{3}-2

6 0
3 years ago
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