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lesya [120]
3 years ago
10

5 divided by a number minus eight.

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
8 0
5/x -8 Let x be the unknown number or variable then that is the equation
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Sonbull [250]
Answer- 3x-56(x) =8
Explanation- Good luck
7 0
4 years ago
If $10,000 is invested at a rate of 2.5 percent for 5 years, how much money is in the account at
Rzqust [24]

Answer:

$8,750

Step-by-step explanation:

Ⓗⓘ ⓣⓗⓔⓡⓔ

Well, first thing's first. Find 2.5% of $10,000.

$10,000*0.025=$250

$250 dollars for 5 years is $250*5=$1,250

Finally, 10,000-$1,250=$8,750

(っ◔◡◔)っ ♥ Hope this helped! Have a great day! :) ♥

Please, please give brainliest, it would be greatly appreciated, I only a few more before I advance, thanks!

6 0
4 years ago
PLEASE HELP I NEED IT TO PASS THIS CLASS
andreyandreev [35.5K]

Answer:

Angle 5 = 37°

Angle 8 = 143°

Step-by-step explanation:

Angle 2 = Angle 8

Reason : Alternate exterior angles are equal.

Therefore,

Angle 8 = 143°

Angle 5 + Angle 8 = 180°

Reason : Angle 5 & 8 are in a linear pair.

Angle 5 + 143° = 180°

Angle 5 = ( 180 - 143 )°

Angle 5 = 37°

4 0
3 years ago
Convert to an improper fraction. Type in your answer with the negative in the numerator.
erastova [34]
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5 0
3 years ago
Evaluate the integral ∫2032x2+4dx. Your answer should be in the form kπ, where k is an integer. What is the value of k? (Hint: d
faltersainse [42]

Here is the correct computation of the question;

Evaluate the integral :

\int\limits^2_0 \ \dfrac{32}{x^2 +4}  \ dx

Your answer should be in the form kπ, where k is an integer. What is the value of k?

(Hint:  \dfrac{d \ arc \ tan (x)}{dx} =\dfrac{1}{x^2 + 1})

k = 4

(b) Now, lets evaluate the same integral using power series.

f(x) = \dfrac{32}{x^2 +4}

Then, integrate it from 0 to 2, and call it S. S should be an infinite series

What are the first few terms of S?

Answer:

(a) The value of k = 4

(b)

a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}

Step-by-step explanation:

(a)

\int\limits^2_0 \dfrac{32}{x^2 + 4} \ dx

= 32 \int\limits^2_0 \dfrac{1}{x+4}\  dx

=32 (\dfrac{1}{2} \ arctan (\dfrac{x}{2}))^2__0

= 32 ( \dfrac{1}{2} arctan (\dfrac{2}{2})- \dfrac{1}{2} arctan (\dfrac{0}{2}))

= 32 ( \dfrac{1}{2}arctan (1) - \dfrac{1}{2} arctan (0))

= 32 ( \dfrac{1}{2}(\dfrac{\pi}{4})- \dfrac{1}{2}(0))

= 32 (\dfrac{\pi}{8}-0)

= 32 ( (\dfrac{\pi}{8}))

= 4 \pi

The value of k = 4

(b) \dfrac{32}{x^2+4}= 8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -...  \ \ \ \ \ (Taylor\ \ Series)

\int\limits^2_0  \dfrac{32}{x^2+4}= \int\limits^2_0 (8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -...) dx

S = 8 \int\limits^2_0dx - \dfrac{3}{2^1} \int\limits^2_0 x^2 dx +  \dfrac{3}{2^3}\int\limits^2_0 x^4 dx -  \dfrac{3}{2^5}\int\limits^2_0 x^6 dx+ \dfrac{3}{2^7}\int\limits^2_0 x^8 dx-...

S = 8(x)^2_0 - \dfrac{3}{2^1*3}(x^3)^2_0 +\dfrac{3}{2^3*5}(x^5)^2_0- \dfrac{3}{2^5*7}(x^7)^2_0+ \dfrac{3}{2^7*9}(x^9)^2_0-...

S= 8(2-0)-\dfrac{1}{2^1}(2^3-0^3)+\dfrac{3}{2^3*5}(2^5-0^5)- \dfrac{3}{2^5*7}(2^7-0^7)+\dfrac{3}{2^7*9}(2^9-0^9)-...

S= 8(2-0)-\dfrac{1}{2^1}(2^3)+\dfrac{3}{2^3*5}(2^5)- \dfrac{3}{2^5*7}(2^7)+\dfrac{3}{2^7*9}(2^9)-...

S = 16-2^2+\dfrac{3}{5}(2^2) -\dfrac{3}{7}(2^2)  + \dfrac{3}{9}(2^2) -...

S = 16-4 + \dfrac{12}{5}- \dfrac{12}{7}+ \dfrac{12}{9}-...

a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}

6 0
3 years ago
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