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2 years ago

The volume of the above cylinder is 21m find the radius of the circular end

Mathematics

1 answer:

PilotLPTM [1.2K]2 years ago

8 0

Step-by-step explanation:

hmmm all of u stay safe

4301154259

Pas 1234

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Which is a better buy?<br><br> 22 vitamins for $2 or 40 vitamins for $4

fomenos

Answer:

i would say 40 for 4$

Step-by-step explanation:

bc ur getting double amount for 2$ and it will last longer

5 0

2 years ago

Read 2 more answers

$1400 is 7% of what number?<br><br> pls help !! ❤️❤️

Scilla [17]

Answer:

$20,000

Explanation:

1400 × 100 ÷ 7

= 20,000

3 0

3 years ago

Assuming that the equation defines x and y implicitly as differentiable functions xequalsf(t), yequalsg(t), find the slope of

Doss [256]

Answer:

$\dfrac{dx}{dt} = -8,\dfrac{dy}{dt} = 1/8\\$

Hence, the slope , $\dfrac{dy}{dx} = \dfrac{-1}{64}$

Step-by-step explanation:

We need to find the slope, i.e. $\dfrac{dy}{dx}$ .

and all the functions are in terms of $t$ .

So this looks like a job for the 'chain rule', we can write:

$\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx} -Eq(A)$

Given the functions

$x = f(t)\\y = g(t)\\$

and

$x^3 +4t^2 = 37 -Eq(B)\\2y^3 - 2t^2 = 110 - Eq(C)$

we can differentiate them both w.r.t to $t$

first we'll derivate Eq(B) to find dx/dt

$x^3 +4t^2 = 37\\3x^2\frac{dx}{dt} + 8t = 0\\\dfrac{dx}{dt} = \dfrac{-8t}{3x^2}\\$

we can also rearrange Eq(B) to find x in terms of t , $x = (37 - 4t^2)^{1/3}$ . This is done so that $\frac{dx}{dt}$ is only in terms of t.

$\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\$

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

$\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\\dfrac{dx}{dt} = \dfrac{-8(3)}{3(37 - 4(3)^2)^{2/3}}\\\dfrac{dx}{dt} = -8$

now let's differentiate Eq(C) to find dy/dt

$2y^3 - 2t^2 = 110\\6y^2\frac{dy}{dt} -4t = 0\\\dfrac{dy}{dt} = \dfrac{4t}{6y^2}$

rearrange Eq(C), to find y in terms of t, that is $y = \left(\dfrac{110 + 2t^2}{2}\right)^{1/3}$ . This is done so that we can replace y in $\frac{dy}{dt}$ to make only in terms of t

$\dfrac{dy}{dt} = \dfrac{4t}{6y^2}\\\dfrac{dy}{dt}=\dfrac{4t}{6\left(\dfrac{110 + 2t^2}{2}\right)^{2/3}}\\$

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

$\dfrac{dy}{dt} = \dfrac{4(3)}{6\left(\dfrac{110 + 2(3)^2}{2}\right)^{2/3}}\\\dfrac{dy}{dt} = \dfrac{1}{8}$

Finally we can plug all of our values in Eq(A)

but remember when plugging in the values that $\frac{dy}{dt}$ is being multiplied with $\frac{dt}{dx}$ and NOT $\frac{dx}{dt}$ , so we have to use the reciprocal!

$\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx}\\\dfrac{dy}{dx} = \dfrac{1}{8}.\dfrac{1}{-8} \\\dfrac{dy}{dx} = \dfrac{-1}{64}$

our slope is equal to $\dfrac{-1}{64}$

7 0

3 years ago

What is the completely factored form of f(x) = x^3-7x^2 2x +4?

Roman55 [17]

The answer is (x-1) (x^2-6x-4)

5 0

2 years ago

Please could someone explain how to do question 2)a) (ii)

KonstantinChe [14]

Answer:

15 kilometres

Step-by-step explanation:

Based on the wording of your question, I am assuming you already know that Paul ran 250 metres every 1 minute, and that the ratio is 250:1. To get the second part of the answer, convert metres to kilometres. 0.001*250= 0.25. Paul ran 0.25 kilometres per minute. The ratio of kilometres per minute is 0.25:1. Next, convert minutes to hours. Multiply the entire ratio by 60, as there are 60 minutes in an hour. The ratio of kilometeres per minute is 15:60, and the ratio of kilometres per hour is 15:1, meaning Paul ran 15 kilometres in one hour.

8 0

2 years ago