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Snowcat [4.5K]
3 years ago
8

Suppose you are a salesperson for Quark Computer Company. Each month you earn $500 plus one sixth of your sales. What amount mus

t you sell this month to earn $3000?​
Mathematics
1 answer:
Yuliya22 [10]3 years ago
5 0

Answer:

sales equal $15,000

Step-by-step explanation:

x = amount of sales

f(x) = 500 + 1/6x

3000 = 500 = x/6

2500 = x/6

x = 15,000

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3 years ago
Use induction to show that 12 + 22 + 32 + ... + n2 = n(n+1)(2n+1)/6, for all n > 1.
dlinn [17]

Answer with Step-by-step explanation:

Let P(n)=1^2+2^2+3^2+.....+n^2=\frac{n(n+1)(2n+1)}{6}

Substitute n=2

Then  P(2)=1+2^2=5

P(2)=\frac{2(2+1)(4+1)}{6}=5

Hence, P(n) is true for n=2

Suppose that P(n) is true for n=k >1

P(k)=1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}

Now, we shall prove that p(n) is true for n=k+1

P(k+1)=1^2+2^2+3^2+...+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}

LHS

P(k+1)=1^2+2^2+3^2+.....+k^2+(k+1)^2

Substitute the value of P(k)

P(k+)=\frac{k(k+1)(2k+1)}{6}+(k+1)^2

P(k+1)=(k+1)(\frac{k(2k+1}{6})+k+1)

P(k+1)=(k+1)(\frac{2k^2+k+6k+6}{6})

P(k+1)=(k+1)(\frac{2k^2+7k+6}{6})

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P(k+1)=\frac{(k+1)(k+2)(2k+3)}{6}

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Hence, P(n) is true for all n >1.

Hence, proved

4 0
3 years ago
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Note that ∠ Q and ∠ R are congruent, that is both 30°, thus

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6 0
4 years ago
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LekaFEV [45]
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