What is the answer to the problem? 7 1/2 ÷ 3/4 I need this fast please help!! :)

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 sum of the polynomials?<br><br> (7x^3 – 4x^2) + (2x^3 – 4x^2)

AVprozaik [17]

(7x^3 – 4x^2) + (2x^3 – 4x^2)
Add 2x^3 to 7x^3
(9x^3 - 4x^2) + (-4x^2)
Add -4x^2 to -4x^2. This would be the same as subtracting 4x^2 from -4x^2
Final Answer: 9x^3 - 8x^2

*The answer cannot be broken down anymore. It's in it's simplest form.

6 0

3 years ago

Someone answer this quickly

san4es73 [151]

The answer is A because it beeds to go over the Y axis

3 0

2 years ago

Read 2 more answers

Determine the equation of straight line which passes through the point (a, b)and perpendicular to ax+by=5

Citrus2011 [14]

Answer:

$y=\frac{b}{a} x.$

Step-by-step explanation:

1) if the required straight line is 'l', the given point is A(a;b), then the required equation of line can be written in form:

$\frac{x-a}{x_1-a} =\frac{y-b}{y_1-b},$

where (x₁;y₁) is other point B, which belongs to the 'l';

2) from the equation it is possible to detect the coordinates of the perpendicular, they are (x₁-a;y₁-b);

3) if the given perpendicular is ax+by=5, then the coordinates of the vector (a;b) are coordinates of the vector, which belongs to the required line 'l', and then: x₁-a=a and y₁-b=b;

4) if to substitute the a=x₁-a and b=y₁-b into the required equation of line 'l', then:

$\frac{x-a}{a} =\frac{y-b}{b};$

5) finally, the equation is: ay=bx, or y=b/a *x (slope-interception form).

note: the suggested solution is not the only way.

5 0

3 years ago

Write a real-world situation that could be modeled by the equation <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B24%7D%7By%7D%2

Alisiya [41]

It whould be 16 to the nearest sixth

7 0

3 years ago