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andriy [413]
3 years ago
15

Find the square root of (3√5-3)³

Mathematics
1 answer:
lys-0071 [83]3 years ago
6 0

Answer:

50.653

Step-by-step explanation:

\sqrt{5 }  = 2.23

3 \times 2.23

6.69

6.69 - 3 = 3.69

{3.69}^{3}

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What are the zeros of the polynomial function f(x)=x3-7x2+8x+16
labwork [276]

Answer: x=4, -1

Step-by-step explanation:

Assuming you meant x^3-7x^2+8x+16, the zeros of the question are x = 4 and -1.

Step 1. Replace f(x) with y.

y = x^3-7x^2+8x+16

Step 2. To find the roots of the equation, replace <em>y</em> with 0 and solve.

0 = x^3-7x^2+8x+16

Step 3. Factor the left side of the equation.

(x-4)^2 (x+1)=0

Step 4. Set x-4 equal to 0 and solve for <em>x</em>.

x-4=0

Step 5. Set x+1 equal to 0 and solve for <em>x</em>.

x=-1

The solution is the result of x-4=0 and x+1=0.

x=4,-1

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3 years ago
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Step-by-step explanation:

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And that the terms of this series may be arranged without changing the value of the series, determine the sum of the reciprocals
nignag [31]

The terms of this series may be arranged without changing the value of the series. The sum of the reciprocals of the squares of the odd positive integers is \pi ^{2} /8.

In mathematics, a sequence is the cumulative sum of a given collection of terms. Usually, those phrases are actual or complicated numbers, but plenty of extra generalities are feasible.

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k=1

1/(1)2+1/(2)2+1/(3)2+1/(4)2+1/(5)2+1/(6)2+1/(7)2+.

up to ∞ terms = 2/6

[1/(1)2+1/(3)2+1/(5)2+1/(7)2+]+[1/(2)²+1/(4)²+1/(6)²+

..∞0] = T²/6

→ [1/(1)² + 1/(3)² + 1/(5)2+1/(7)2+......00] + [1/4 (1)² + 1/4(2)²+

1/4(3)²+....0] =²/6

[1/(1)²+1/(3)²+1/(5)2+1/(7)2+.......)] + 1/4[1/(1)² + 1/(2)²+

1/(3)²+....x] = 2/6

⇒ [1/(1)² + 1/(3)² + 1/(5)²+1/(7)²+..] + 1/4 [π²/6] = 2/6

⇒ [1/(1)² + 1/(3)² + 1/(5)²+1/(7)²+] = (1-1/4)/6

⇒ [1/(1)²+1/(3)2+1/(5)2+1/(7)2+..∞ = 3/4 x π²/6

=

↑

[1/(1)2+1/(3)2+1/(5)2+1/(7)²+] = 2/8

Learn  more about the series here brainly.com/question/24295771

#SPJ4

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