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Stels [109]
3 years ago
12

A store charges $20 for 15 candy bars. at this rate how much would they charge for 18 candy bars.

Mathematics
2 answers:
sammy [17]3 years ago
4 0

23 im pretty sure lol

Step-by-step explanation: If not im sorry i couldn't help and have a good day lol

Alecsey [184]3 years ago
3 0
$360 Keel multiplying 20 By 18
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larisa [96]

Answer:

512

Step-by-step explanation:

8 x 8 x 8 = 512

4 0
3 years ago
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What is 950% as a fraction
mrs_skeptik [129]

Answer:

9 1/2

Step-by-step explanation:

\frac{950}{100}=9\frac{50}{100} = 9 \frac{1}{2}

3 0
3 years ago
The pirce of an item yesterday was$125 . Today, the price fell to 75. Find the percentage decrease.
Julli [10]

Answer:

Im sure the answer is 40%

<em>GL Deary</em>

<em>Hope I helped!!!</em>

<em><3</em>

8 0
4 years ago
Can you prove it??<br>it's hard,I tried but couldn't solve this.
BARSIC [14]

I'll abbreviate s=\sin\theta and c=\cos\theta, so the identity to prove is

\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1s\right)

On the left side, we can simplify a bit:

\dfrac{s+c+1}{s+c-1}=\dfrac{s+c-1+2}{s+c-1}=1+\dfrac2{s+c-1}

\dfrac{1+s-c}{1-s+c}=-\dfrac{-2+1-s+c}{1-s+c}=-1+\dfrac2{1-s+c}

Then

\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1{s+c-1}-\dfrac1{1-s+c}\right)

So the establish the original equality, we need to show that

\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac1s

Combine the fractions:

\dfrac{(1-s+c)-(s+c-1)}{(s+c-1)(1-s+c)}=\dfrac{2-2s}{c^2-s^2+2s-1}

We can rewrite the denominator as

c^2-s^2+2s-1=c^2+s^2-2s^2+2s-1

then using the fact that c^2+s^2=\cos^2\theta+\sin^2\theta=1, we get

1-2s^2+2s-1=2s-2s^2

so that we have

\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac{2-2s}{2s-2s^2}=\dfrac1s

as desired.

6 0
4 years ago
Harper is a salesperson who sells computers at an electronics store. She makes a base pay amount of $100 per day regardless of s
Studentka2010 [4]

Answer:

P=0.01x+100

Step-by-step explanation:

3 0
3 years ago
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