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Nikolay [14]
3 years ago
12

Find five consecutive integers such that the sum of the first and 5 times the third is equal to 41 less than 3 times the sum of

the second fourth and fifth
Mathematics
1 answer:
bogdanovich [222]3 years ago
6 0

Answer:

see below

Step-by-step explanation:

We'll cal the first integer x and then the rest of them will be x + 1, x + 2, x + 3 and x + 4. We can write x + 5(x + 2) = 3(x + 1 + x + 3 + x + 4) - 41.

x + 5x + 10 = 3(3x + 8) - 41

6x + 10 = 9x + 24 - 41

6x + 10 = 9x - 17

3x = 27

x = 9

The numbers are 9, 10, 11, 12, 13.

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In Problems 23–30, use the given zero to find the remaining zeros of each function
Talja [164]

Answer:

x =  2i, x = -2i and x = 4 are the roots of given polynomial.

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We are given the following expression in the question:

f(x) = x^3 - 4x^2+ 4x - 16

One of the zeroes of the above polynomial is 2i, that is :

f(x) = x^3 - 4x^2+ 4x - 16\\f(2i) = (2i)^3 - 4(2i)^2+ 4(2i) - 16\\= -8i+ 16+8i-16 = 0

Thus, we can write

(x-2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

Now, we check if -2i is a root of the given polynomial:

f(x) = x^3 - 4x^2+ 4x - 16\\f(-2i) = (-2i)^3 - 4(-2i)^2+ 4(-2i) - 16\\= 8i+ 16-8i-16 = 0

Thus, we can write

(x+2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

Therefore,

(x-2i)(x+2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16\\(x^2 + 4)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

Dividing the given polynomial:

\displaystyle\frac{x^3 - 4x^2 + 4x - 16}{x^2+4} = x -4

Thus,

(x-4)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

X = 4 is a root of the given polynomial.

f(x) = x^3 - 4x^2+ 4x - 16\\f(4) = (4)^3 - 4(4)^2+ 4(4) - 16\\= 64-64+16-16 = 0

Thus, 2i, -2i and 4 are the roots of given polynomial.

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3 years ago
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Answer:

9604

Step-by-step explanation:

V = (7/3)³ × 756

= 343/27 × 756

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2 years ago
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