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2 years ago

The two lines represent the amount of water filling over time in two tanks of the same size.

Mathematics

1 answer:

dlinn [17]2 years ago

7 0

Answer:

Tank A is filling quicker

Step-by-step explanation:

The slope of line A is larger and going up faster, that tells you tank A is filling up faster than Tank B

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Estimate the sum 258+565

Thepotemich [5.8K]

So do 260+570 which is 830

The real sum is 823

7 0

3 years ago

Integral of x"2+4/x"2+4x+3

dolphi86 [110]

I'm guessing you mean

$\displaystyle \int\frac{x^2+4}{x^2+4x+3}\,\mathrm dx$

First, compute the quotient:

$\displaystyle \frac{x^2+4}{x^2+4x+3} = 1 + \frac{4x-1}{x^2+4x+3}$

Split up the remainder term into partial fractions. Notice that

x ² + 4x + 3 = (x + 3) (x + 1)

Then

$\displaystyle \frac{4x-1}{x^2+4x+3} = \frac a{x+3} + \frac b{x+1} \\\\ \implies 4x - 1 = a(x+1) + b(x+3) = (a+b)x + a+3b \\\\ \implies a+b=4 \text{ and }a+3b = -1 \\\\ \implies a=\frac{13}2\text{ and }b=-\frac52$

So the integral becomes

$\displaystyle \int \left(1 + \frac{13}{2(x+3)} - \frac{5}{2(x+1)}\right) \,\mathrm dx = \boxed{x + \frac{13}2\ln|x+3| - \frac52 \ln|x+1| + C}$

We can simplify the result somewhat:

$\displaystyle x + \frac{13}2\ln|x+3| - \frac52 \ln|x+1| + C \\\\ = x + \frac12 \left(13\ln|x+3| - 5\ln|x+1|\right) + C \\\\ = x + \frac12 \left(\ln\left|(x+3)^{13}\right| - \ln\left|(x+1)^5\right|\right) + C \\\\ = x + \frac12 \ln\left|\frac{(x+3)^{13}}{(x+1)^5}\right| + C \\\\ = \boxed{x + \ln\sqrt{\left|\frac{(x+3)^{13}}{(x+1)^5}\right|} + C}$

3 0

2 years ago

2(4w-1)=-10(w-3)+4 help

solniwko [45]

Answer:

w = 12

Step-by-step explanation:

2(4w-1)=-10(w-3)+4

2*4w + 2*-1 = (10*w + 10*-3) + 4

8w - 2 = 10w - 30 + 4

8w - 2 = 10w - 26

26 - 2 = 10w - 8w

24 = 2w

w = 24/2

w = 12

Check:

2((4*12)-1) = 10(12-3) + 4

2(48-1) = 10(9) + 4

2*47 = 90 + 4 = 94

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3 years ago

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Lerok [7]

Answer:

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3 years ago

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What is the missing length

steposvetlana [31]

Answer:

25 km

Step-by-step explanation:

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3 years ago