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kykrilka [37]
3 years ago
15

Three consecutive odd integers are such that the square of the third integer is 1515 greater than greater than the sum of the sq

uares of the first two. one solution is 33​, 55​, and 77. find three other consecutive odd integers that also satisfy the given conditions.

Mathematics
1 answer:
Ksju [112]3 years ago
7 0
Let
x-----------> first <span>odd integer
x+2--------> second consecutive odd integer
x+4-------> third consecutive odd integer

we know that
(x+4)</span>²=15+x²+(x+2)²-------> x²+8x+16=15+x²+x²+4x+4
x²+8x+16=19+2x²+4x-------> x²-4x+3
x²-4x+3=0

using a graph tool----------> to calculate the quadratic equation
see the attached figure
the solution is
x=1
x=3

the answer is
the first odd integer x is 1
the  second consecutive odd integer x+2 is 3
the third consecutive odd integer x+4 is 5

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Read 2 more answers
Sin tita= 0.6892.find the value Of tita correct to two decimal places<br><br>​
Anvisha [2.4K]

Answer:

\theta \approx 6.28n + 2.38,  \quad  n \in \mathbb{Z}

or

\theta \approx 6.28n + 0.76, \quad n \in \mathbb{Z}

Considering \theta \in (0, 2\pi]

\theta \approx 2.38

or

\theta \approx 0.76

Step-by-step explanation:

\sin(\theta)=0.6892

We have:

\sin (x)=a \Longrightarrow x=\arcsin (a)+2\pi n \text{ or } x=\pi -\arcsin (a)+2\pi n \text{ as } n\in \mathbb{Z}

Therefore,

\theta= \arcsin (0.6892)+2\pi n, \quad n \in \mathbb{Z}

or

\theta = \pi -\arcsin (0.6892)+2\pi n, \quad  n\in \mathbb{Z}

---------------------------------

\theta \approx 6.28n + 2.38,  \quad  n \in \mathbb{Z}

or

\theta \approx 6.28n + 0.76, \quad n \in \mathbb{Z}

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Answer:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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