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1 year ago

Find the indicated limit, if it exists. See file below.

Mathematics

1 answer:

White raven [17]1 year ago

6 0

Answer:

Step-by-step explanation:

The limit exists because it's continuous since they equal each other

when x approaches zero.

Substitute both expressions with x=0 and both of them will equal 2, so the limit should be 2.

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F = uv/u + v make u the subject

never [62]

I'm assuming that you mean

$\dfrac{uv}{u+v}$

because if you meant

$\dfrac{uv}{u}+v$

then u would simplify and you couldn't make it the subject.

Under this assumption, we start with

$F=\dfrac{uv}{u+v}$

We multiply both sides by $u+v$

$F(u+v)=uv$

We expand the left hand side:

$Fu+Fv=uv$

We move all terms involving u to the left and all terms not involving u to the right:

$Fu-uv=-Fv$

We factor u on the left hand side:

$u(F-v)=-Fv$

We divide both sides by $F-v$

$u=\dfrac{-Fv}{F-v}$

3 0

2 years ago

Nandini makes 'halwa' one evening and divides it into four equal portions for her family of four. However, just as they are abou

wel

Answer:

Step-by-step explanation:

Total 'halwa' made = 1

Divided into four equal portion = 1/4

Arrival of an unexpected guest = 1/5

By what percentage has each family member's share been reduced:

Change in the sharing proportion:

Previous share ratio - new sharing ratio

(1/4 - 1/5) = (5 - 4) / 20 = 1/20

That means total reduction in the sharing = 1/ 20

Since each member comes contributed equally:

Reduction in each family member's share ;

(1 / 20) ÷ 4

(1 / 20) * 1/4 = 1/ 80

Percentage reduction:

(Reduction / original share) * 100%

[(1/80) ÷ (1/4)] * 100

(1/80 * 4/1) * 100%

(1/20) * 100%

= 5%

Reduction in each family members share = 5%

3 0

2 years ago

How do you integrate arctan(x dx? i think that if you simplify the integral you get:?

Temka [501]

Integration by parts will help here. Letting $u=\arctan x$ and $\mathrm dv=\mathrm dx$ , you end up with $\mathrm du=\dfrac{\mathrm dx}{1+x^2}$ and $v=x$ . Now

$\displaystyle\int\arctan x\,\mathrm dx=uv-\int v\,\mathrm du$
$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx$

For the remaining integral, setting $y=1+x^2$ gives $\dfrac{\mathrm dy}2=x\,\mathrm dx$ , so

$\displaystyle\int\frac x{1+x^2}\,\mathrm dx=\frac12\int\frac{\mathrm dy}y=\frac12\ln|y|+C=\frac12\ln(1+x^2)+C$

Putting everything together, you end up with

$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C$