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Veronika [31]
2 years ago
11

Calculate the percentage increase for the price of a book increasing from 250 to 275.50​

Mathematics
2 answers:
Naddika [18.5K]2 years ago
6 0

Step-by-step explanation:

increase in price=275.50-250

=75 .50

now,in precent

( increase in price÷initial price )×100%

=(75.50÷250) ×100%

= 30.2%

Vadim26 [7]2 years ago
5 0

Answer:

250×275.50

68,875

answer is 68,875

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Answer:

Let S = starting point

Let ST = length of tunnel

SR = 1414 m

RT = 2236 m

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ST^2 + SR^2 = RT^2

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Step-by-step explanation:

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2 years ago
)) In a right triangle, a and b are the lengths of the legs and c is the length of the
kondor19780726 [428]

Answer:

5.66 inches

Step-by-step explanation:

{a}^{2}  +  {b }^{2}   =   {c}^{2}

a² + 7² = 9²

a² + 49 = 81

a² + 49 - 49 = 81 - 49

a² = 32

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I used the Pythagorean Theorem to solve this.

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3 years ago
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i think the answer is A

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again "i think". so i could be wrong

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Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:
kondaur [170]
\text{Proof by induction:}
\text{Test that the statement holds or n = 1}

LHS = (3 - 2)^{2} = 1
RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS
\text{Thus, the statement holds for the base case.}

\text{Assume the statement holds for some arbitrary term, n= k}
1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}

\text{Prove it is true for n = k + 1}
RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}

LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}
= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}
= \frac{k(6k^{2} + 15k + 11) + 2}{}
= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}
= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}
= RHS

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.
3 0
3 years ago
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