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Bad White [126]
3 years ago
10

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!
You might be interested in
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
Read 2 more answers
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
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