All categories

3 years ago

BRO PLEASE HELP I NEED THIS STUFF IN AN HOUR PLEASE...

Mathematics

1 answer:

inessss [21]3 years ago

5 0

Answer:

They Have The Same Y-Intercept.

Function 2's y-intercept is -5.

Function 1 y-intercept could be found by inputting 0 as r.

p=0-5

p=-5

Meaning, they're equal.

You might be interested in

1. En una biblioteca quieren comprar una computadora que

miss Akunina [59]

Answer:

678.74

that might be the question since I don't really understand what it says really

8 0

3 years ago

Rearrange to make x the subject of 4(x-3)/a=y

IrinaVladis [17]

Answer: x = a*y + 3

Step-by-step explanation:

To make x the subject of the equation, first, we open the bracket

4x - 12/a = y

Then cross multiply:

4x - 12 = a * y ( a*y means the product of the two variables)

Add 12 to both sides of the equation

4x = a*y + 12

Divide both sides by 4 to get the value of x

x = a * y + 12/4

x = a*y + 3

I hope this helps.

4 0

3 years ago

-3/4 is a rational number, but not an integer.<br> True<br> False

lora16 [44]

Answer:

It's true because its a rational number.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

8 0

4 years ago

For a fund raiser, the debate team sold sodas at a school dance. They spent 10 dollars to pay for the cups. The soda was donated

8_murik_8 [283]

Answer:byh

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers