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3 years ago

I really need help would please like help thanks

Mathematics

1 answer:

ASHA 777 [7]3 years ago

4 0

Hey there!

A reciprocal might be an unfamiliar vocab word, but the concept is really quite simple.

It's just a flipped fraction! Multiplying the flipped fraction by it's reciprocal gets you one, and that's how you check your answer.

So we have:

Reciprocal of 9/10 is 10/9, because 9/10 times 10/9 = 90/90 = 1

Reciprocal of 1/11 is 11/1, because 1/11 times 11/1 = 11/11 = 1

Reciprocal of 10(remember, it's 10/1) is 1/10, because 10/1 times 1/10 = 10/10 = 1

Hope this helps!

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Make a substitution to express the integrand as a rational function and then evaluate the integral. (Use C for the constant of i

saw5 [17]

Answer:

$I = - 7 \cdot \ln|e^{x}+7| + 9 \cdot \ln|e^{x}+9| + C$

Step-by-step explanation:

Let consider the following algebraical substitution:

$u = e^{x}$

$du = e^{x} dx$

The integral is re-arranged to this form:

$I = 2 \cdot \int\ {\frac{u\ du}{u^{2}+16\cdot u + 63} } \, dx$

After doing some algebraical manipulation, the rational form of this integral is:

$I = -\int {\frac{7}{u + 7} } \, dx + \int {\frac{9}{u + 9} } \, dx \\$

The solution of this integral is:

$I = - 7 \cdot \ln|u+7| + 9 \cdot \ln|u+9| + C$

$I = - 7 \cdot \ln|e^{x}+7| + 9 \cdot \ln|e^{x}+9| + C$

6 0

4 years ago

find a curve that passes through the point (1,-2 ) and has an arc length on the interval 2 6 given by 1 144 x^-6

taurus [48]

Answer:

$f(x) = \frac{6}{x^2} -8$ or $f(x) = -\frac{6}{x^2} + 4$

Step-by-step explanation:

Given

$(x,y) = (1,-2)$ --- Point

$\int\limits^6_2 {(1 + 144x^{-6})} \, dx$

The arc length of a function on interval [a,b]: $\int\limits^b_a {(1 + f'(x^2))} \, dx$

By comparison:

$f'(x)^2 = 144x^{-6}$

$f'(x)^2 = \frac{144}{x^6}$

Take square root of both sides

$f'(x) =\± \sqrt{\frac{144}{x^6}}$

$f'(x) = \±\frac{12}{x^3}$

Split:

$f'(x) = \frac{12}{x^3}$ or $f'(x) = -\frac{12}{x^3}$

To solve fo f(x), we make use of:

$f(x) = \int {f'(x) } \, dx$

For: $f'(x) = \frac{12}{x^3}$

$f(x) = \int {\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = \frac{12}{2x^2} + c$

$f(x) = \frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = \frac{6}{1^2} + c$

$-2 = \frac{6}{1} + c$

$-2 = 6 + c$

Make c the subject

$c = -2-6$

$c = -8$

$f(x) = \frac{6}{x^2} + c$ becomes

$f(x) = \frac{6}{x^2} -8$

For: $f'(x) = -\frac{12}{x^3}$

$f(x) = \int {-\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = -\frac{12}{2x^2} + c$

$f(x) = -\frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = -\frac{6}{1^2} + c$

$-2 = -\frac{6}{1} + c$

$-2 = -6 + c$

Make c the subject

$c = -2+6$

$c = 4$

$f(x) = -\frac{6}{x^2} + c$ becomes

$f(x) = -\frac{6}{x^2} + 4$

3 0

3 years ago

guys I need help with this problem. I've tried solving it but everything I do is wrong please help me and explain what you did.

Vaselesa [24]

Answer:

This is incorrect because when you seperate both equations, the 7/x flips to x/7. x times x does not cancel out, and the correct division gives you x²/21

Step-by-step explanation:

(x/3)/(7/x)

(x/3) ÷ (7/x)

(x/3) × x/7

x²/21

6 0

3 years ago

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If points c, d, and e are on a line and cd =20 and ce = 32 what are the possible values of de

vivado [14]

In this question it is given that the values of CD and CE are 20 and 32 respectively.

Since the length of CE is greater then of CD, so C and D could not be the end points.

Therefore, either D and E are the endpoints or C and E are the endpoints.

If D and E are the endpoints, then DE is the sum of 20 and 32 which is 52.

And if C and E are the endpoints, then DE is the difference of CE and CD that is 12.

So the possible values of DE are 12 and 52.

6 0

3 years ago

What is the volume of a cylinder, in cubic centimeters, with a height of 18 centimeters and a base diameter of 8 centimeters? Ro

marissa [1.9K]

Answer:

V≈904.78 = 900

hope this helps :) plz brainliest

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers