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irina1246 [14]
2 years ago
11

Make v the subject of the formula in H=m(v²-u²)/2gx​

Mathematics
1 answer:
goldfiish [28.3K]2 years ago
4 0

Answer:

v=\sqrt{\dfrac{H\times 2gx}{m}+u^2}

Step-by-step explanation:

The given formula is :

H=\dfrac{m(v^2-u^2)}{2gx}

We need to find the formula for v.

Cross multiplying the given formula.

H\times 2gx=m(v^2-u^2)

Dividing both sides by m. So,

\dfrac{H\times 2gx}{m}=\dfrac{m(v^2-u^2)}{m}\\\\(v^2-u^2)=\dfrac{H\times 2gx}{m}

Adding both sides by u².

(v^2-u^2)+u^2=\dfrac{H\times 2gx}{m}+u^2\\\\v^2=\dfrac{H\times 2gx}{m}+u^2\\\\\text{So},\\\\v=\sqrt{\dfrac{H\times 2gx}{m}+u^2}

Hence, this is the required formula.

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If the limit of f(x) as x approaches 8 is 3, can you conclude anything about f(8)?
natali 33 [55]

If the limit of f(x) as x approaches 8 is 3, can you conclude anything about f(8)? The answer is No. We cannot. See the explanation below.

<h3>What is the justification for the above position?</h3>

Again, 'No,' is the response to this question. The justification for this is that the value of a function does not depend on the function's limit at a given moment.

This is particularly clear when we consider a question with a gap. A rational function with a hole is an excellent example that will help you answer this question.

The limit of a function at a position where there is a hole in the function will exist, but the value of the function will not.

<h3>What is limit in Math?</h3>

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6 0
1 year ago
Find the taylor series for f(x) centered at the given value of a. [assume that f has a power series expansion. do not show that
RUDIKE [14]

The taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 (x+4)^{2}/2!+...........

Given a function f(x)=9/x,a=-4.

We are required to find the taylor series for the function f(x)=8/x centered at the given value of a and a=-4.

The taylor series of a function f(x)=f(a)+f^{1}(a)(x-a)/1!+ f^{11}(a)(x-a)^{2} /2! +f^{111}(a)(x-a)a^{3}/3!+..........

Where the terms in f prime f^{1}(a) represent the derivatives of x valued at a.

For the given function.f(x)=8/x and a=-4.

So,f(a)=f(-4)=8/(-4)=-2.

f^{1}(a)=f^{1}(-4)=-8/(-4)^{2}

=-8/16

=-1/2

The series of f(x) is as under:

f(x)=f(-4)+f^{1}(-4)(x+4)/1!+  f^{11}(-4)(x+4)^{2}/2!.............

=8/(-4)-8/(-4)^{2} (-4)(x+4)/1!+  24/(-4)^{3} (-4)(x+4)^{2}/2!.............

=-2+2(x+4)/1!-24/16 (x+4)^{2}/2!+...........

Hence the taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 (x+4)^{2}/2!+...........

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3 0
1 year ago
Solve: 5 - 3x &lt; 20<br> Help
Fed [463]

Answer:

hope this helps

Step-by-step explanation:

x > −5

7 0
3 years ago
Read 2 more answers
Write 3,200 in scientific notation
spin [16.1K]

Answer:

The answer is 3.2 E 3 or 3.2 x 10^3

You move back 3 spaces which is 3.2, since you moved back 3 spaces, you put 3 as the power/exponet.

To be more satisfied, 3.2 is your base number, it is the big number Below the power/exponent. The power/exponent is 3, you moved 3 spaces and you moved to the left, so it is represented as a positive power/exponent.

Hope this helps.

8 0
2 years ago
HELP PLEASE <br>Use greater than, equal to =, or less than sign to compare<br> 2/3 and 1/6
lana [24]

Answer:

2/3 < 1/6

Step-by-step explanation:

hope this helps

3 0
3 years ago
Read 2 more answers
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