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3 years ago

Four friends charge different amounts for babysitting. Who charges the most per hour?

Mathematics

2 answers:

lesya [120]3 years ago

8 0

We need the rates please attach a picture

dalvyx [7]3 years ago

3 0

Answer:

Xiavier changes the most :).

Step-by-step explanation:

Claudia work for 3 hours to get $28

Xiavier works for 2 hours to get $20

Ling works for 4 hours to get $30

And Alyssa works 6 hours to get $45

Claudia is charging $9.33 per hour

Xiavier is charging $10 per hour

Ling is charging $7.50 per hour

and Alyssa is charging 7.50 per hour.

This makes your answer Xiavier.

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What is <br>60,845-2,926=?

Alchen [17]

Answer:

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Step-by-step explanation:

60,845-2,926=57,919

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zhenek [66]

Hi!

The GCM of 14 and 28 is 7. 7 goes into 14 2 times, and goes into 28 4 times. So...

2/4

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Anybody help me to solve this question.

Mumz [18]

Answer:

$\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)} are\ in\ AP$

Step-by-step explanation:

Given that $(b-c)^2, (c-a)^2 , (a-b)^2$ are in AP

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From given as we know if p , q, r are in AP then 2q= p+r.

$2(c-a)^2= (b-c)^2+(a-b)^2\\\\\Rightarrow 2(c^2+a^2-2ac)=b^2+c^2-2bc+a^2+b^2-2ab\\\\\Rightarrow 2c^2+2a^2-4ac= 2b^2+c^2+a^2 -2bc-2ab\\\\\Rightarrow a^2+c^2-2b^2-4ac= -2bc-2ab\\\\\Rightarrow a^2-2b^2+c^2= 4ac-2bc-2ab$

Now

$\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)}2\dfrac{1}{(c-a)} =\dfrac{1}{(b-c)}+\dfrac{1}{(a-b)}\\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-b+b-c}{(b-c)(a-b)} \\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-c}{(b-c)(a-b)} \\\\\Rightarrow2(b-c)(a-b) = (c-a)(a-c) \\\\\Rightarrow 2(ab-b^2-ac+bc)= -(a-c)^2\\\\\Rightarrow 2ab- 2b^2-2ac+2bc = -a^2-c^2+2ac\\\\\Rightarrow a^2-2b^2+c^2=4ac-2ab-2bc$

Which is the result of AP

Hence proved

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3 years ago

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Yuki888 [10]

Answer:

i got 7

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