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Paul [167]
3 years ago
13

Perry made $20,000 per year.He soon he received a promotion(raise) with a 20% increase in pay.What is his new salary?How much is

the promotion
Mathematics
1 answer:
user100 [1]3 years ago
5 0
Hello! The salary goes up 20% due to the promotion. His original salary was $20,000. To find out his new salary, you have to do is multiply the original price by the rate. 20,000 * 20% (0.2) is 4,000. His promotion is $4,000. Now, we add. 20,000 + 4,000 is 24,000. There. Perry's new salary is $24,000.
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Six less than the product of five and a number is ten.
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Answer:

• Let the number be x

{ \boxed{ \boxed{ \boxed{ \mathfrak{ \: answer \: }} \:  \dashrightarrow \: } \: { \rm{5x - 6 = 10 \: }}}}

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The relation Q is described as a list of ordered pairs, shown below.
Finger [1]

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Domain : {-2, -1, 0, 4}

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Let x,y \epsilon R. Use mathmatical induction to prove the identity.
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Step-by-step explanation:

We will prove by mathematical induction that, for every natural n,  

(x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})=x^{n+1}-y^{n+1}

We will prove our base case (when n=1) to be true:

Base case:

(x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})=(x-y)(x^{1}+y^{1})=x^2-y^2=x^{1+1}-y^{1+1}

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(x-y)(x^{n+1}+x^{n}y+...+xy^{n}+y^{n+1})=(x-y)y(\frac{x^{n+1}}{y}+x^{n}+...+xy^{n-1}+y^{n})=(x-y)y(\frac{x^{n+1}}{y})+(x-y)y(x^{n}+...+xy^{n-1}+y^{n})=(x-y)y(\frac{x^{n+1}}{y})+y(x^{n+1}-y^{n+1})=(x-y)x^{n+1}+y(x^{n+1}-y^{n+1})=x^{n+2}-yx^{n+1}+yx^{n+1}-y^{n+2}=x^{n+2}-y^{n+2}\\

With this we have proved our statement to be true for n+1.    

In conlusion, for every natural n,

(x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})=x^{n+1}-y^{n+1}

8 0
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