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evablogger [386]
4 years ago
8

PLZZ HELP this is overdue, and my last question ::>_<::

Mathematics
2 answers:
Virty [35]4 years ago
8 0

Answer:

hmmmm lemme think. id say simplify the top and bottom then divide. im kinda busy rn so i cant do it. sorry bro but what grade is this and i could help you some more in the future ok :)

Step-by-step explanation:

skad [1K]4 years ago
6 0
A+3 this the answer I hoped I helped
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The circumference of a circle is 60pi cm. What is the radius of the circle ?
Margaret [11]

Answer:

30 cm

Step-by-step explanation:

The circumference of a circle is

C = 2*pi*r

60pi = 2* pi *r

Divide each side by 2pi

60/2pi = 2 pi r/ 2pi

30 =r

5 0
4 years ago
Read 2 more answers
What is the measure of c?<br> a = 15<br> j = 24<br> k = 32
inessss [21]

Answer:

The answer to your question is c = 20

Step-by-step explanation:

Data

a = 15

j = 24

k = 32

c = ?

Process

Use proportions to solve this problem. Compare the small triangle and the large triangle.

1.- Proportion of the small triangle

             c/15

2.- Proportion of the large triangle

            k/j

3.- Equal but terms and solve for c

            c/15 = k/j

-Substitution

            c / 15 = 32/24

            c = 32(15)/24

-Result

            c = 20

       

4 0
4 years ago
What is the missing constant term in the perfect square that starts with x^2+6x ?
Olegator [25]
X^2 + 6x + 9
x                3
x                3

(x+3) (x+3)

x^2 + 3x + 3x + 9         (FOIL method)
x^2 + 6x + 9 (true)

your answer is 9 as the constant

hope this helps
8 0
4 years ago
Read 2 more answers
What is the interquartile range of the data set<br><br> {76, 22, 38, 95, 75, 60, 62, 92}
leonid [27]
The answer is 35. To find it, put the numbers in chronological order, the split the set in half. Find the middle of the first half, which is 49. Then, find the middle of the second half, which is 84. Lastly, subtract 49 from 84 to get 35! Hope that helps!
4 0
4 years ago
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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