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Dvinal [7]
3 years ago
10

Solve this please.

3(m - 5) + 11 = - 85" alt="3(m - 5) + 11 = - 85" align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
Vikki [24]3 years ago
7 0

Answer:

Bro this some alien hyroglyphics

Step-by-step explanation:

Marrrta [24]3 years ago
6 0

Answer:

m = - 27

Step-by-step explanation:

Given

3(m - 5) + 11 = - 85 ← distribute and simplify left side

3m - 15 + 11 = - 85

3m - 4 = - 85 ( add 4 to both sides )

3m = - 81 ( divide both sides by 3 )

m = - 27

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Help me to simplify fully
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Read 2 more answers
How do you do this question?
lukranit [14]

Answer:

0.001591

Step-by-step explanation:

The power series for arctan(x) is:

arctan(x) = ∑ (-1)ⁿ x²ⁿ⁺¹ / (2n + 1)

Substituting 5x:

arctan(5x) = ∑ (-1)ⁿ (5x)²ⁿ⁺¹ / (2n + 1)

Multiply both sides by x:

x arctan(5x) = ∑ (-1)ⁿ x (5x)²ⁿ⁺¹ / (2n + 1)

Simplify:

x arctan(5x) = ∑ (-1)ⁿ (5x) (5x)²ⁿ⁺¹ / (10n + 5)

x arctan(5x) = ∑ (-1)ⁿ (5x)²ⁿ⁺² / (10n + 5)

Multiply top and bottom by 5:

x arctan(5x) = ∑ (-1)ⁿ 5 (5x)²ⁿ⁺² / (50n + 25)

Integrate:

∫ x arctan(5x) = ∑ (-1)ⁿ (5x)²ⁿ⁺³ / ((50n + 25) (2n + 3))

Evaluate between x = 0.1 and x = 0:

∫₀⁰¹ x arctan(5x) = [∑ (-1)ⁿ (5x)²ⁿ⁺³ / ((50n + 25) (2n + 3))] |₀⁰¹

∫₀⁰¹ x arctan(5x) = ∑ (-1)ⁿ (0.5)²ⁿ⁺³ / ((50n + 25) (2n + 3))

This is an alternating series.  We can approximate it with Alternating Series Estimation.

bₙ₊₁ ≥ ε

(0.5)²⁽ⁿ⁺¹⁾⁺³ / ((50(n+1) + 25) (2(n+1) + 3)) ≥ 0.000001

(0.5)²ⁿ⁺⁵ / ((50n + 75) (2n + 5)) ≥ 0.000001

n ≥ 3

So the approximation is the sum of the terms from n=0 to n=3.

(-1)⁰ (0.5)²⁽⁰⁾⁺³ / ((50(0) + 25) (2(0) + 3))

+ (-1)¹ (0.5)²⁽¹⁾⁺³ / ((50(1) + 25) (2(1) + 3))

+ (-1)² (0.5)²⁽²⁾⁺³ / ((50(2) + 25) (2(2) + 3))

+ (-1)³ (0.5)²⁽³⁾⁺³ / ((50(3) + 25) (2(3) + 3))

= 0.0016667 − 0.0000833 + 0.0000089 − 0.0000012

= 0.001591

7 0
3 years ago
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