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3 years ago

What number do I subtract each time?

Mathematics

2 answers:

alexgriva [62]3 years ago

5 0

Step-by-step explanation:

as you can see in-between 345 and 350 are 5 spaces. 5 plus 345 equals 350.

tiny-mole [99]3 years ago

4 0

Subtract 1 because the number is going down by one

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Simplify using the product rule.<br> a^4 x a^3

cluponka [151]

Answer: the answer to the question is a7x

Step-by-step explanation:

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3 years ago

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3 years ago

Given A = {x | x < 1}, B = {x | x ≥ 5}, and C = {x | x = 5} Select all that are true:

ladessa [460]

Answer:

A ∩ B = The empty set

A U C = {x│x < 1 or x = 5}

B ∩ C = {x│x=5}

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3 years ago

the art club held a show for 2 days a total 269 people attended the show. On the second day, 15 more people attended than had co

zmey [24]

Answer:

127 people

Step-by-step explanation:

Number of attendees the first day = x.

<u>Solve the following equation for x:</u>

x + (x + 15) = 269
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5 0

3 years ago

Calculus 2 Master needed, show steps with partial fraction decomposition <img src="https://tex.z-dn.net/?f=%5Cint%283x%5E2-26x%2

Vladimir79 [104]

Answer:

-ln|x−5| + 2 ln(x²+4) + 3 tan⁻¹(x/2) + C

Step-by-step explanation:

The fraction will be split into a sum of two other fractions.

The first fraction will have a denominator of x − 5. The numerator will the a polynomial of one less order, in this case, a constant A.

The second fraction will have a denominator of x² + 4. The numerator will be Bx + C.

$\frac{3x^{2}-26x+26}{(x-5)(x^{2}+4)}=\frac{A}{x-5} +\frac{Bx+C}{x^{2}+4}$

Combine the two fractions back into one using the common denominator.

$\frac{A}{x-5} +\frac{Bx+C}{x^{2}+4}=\frac{A(x^{2}+4)+(Bx+C)(x-5)}{(x-5)(x^{2}+4)}$

This numerator will equal the original numerator.

$A(x^{2}+4)+(Bx+C)(x-5)=3x^{2}-26x+26\\Ax^{2}+4A+Bx^{2}-5Bx+Cx-5C=3x^{2}-26x+26\\(A+B)x^{2}+(C-5B)x+(4A-5C)=3x^{2}-26x+26$

Match the coefficients.

$A+B=3\\C-5B=-26\\4A-5C=26$

Solve the system of equations.

$A=-1\\B=4\\C=-6$

So we can rewrite the integral as:

$\int {(\frac{-1}{x-5}+\frac{4x-6}{x^{2}+4}) } \, dx$

Solving:

$\int {\frac{-1}{x-5}\, dx + \int {\frac{4x}{x^{2}+4}} \, dx - \int {\frac{6}{x^{2}+4} } \, dx$

$-\int {\frac{1}{x-5}\, dx + 2\int {\frac{2x}{x^{2}+4}} \, dx - 6\int {\frac{1}{x^{2}+4} } \, dx$

$-ln|x-5| + 2ln(x^{2}+4) - 6(\frac{1}{2} tan^{-1}(\frac{x}{2} )) + C$

$-ln|x-5| + 2ln(x^{2}+4) - 3 tan^{-1}(\frac{x}{2} ) + C$

5 0

3 years ago