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2 years ago

10

Can someone please give me the (Answers) to this? ... please ...

1 answer:

Igoryamba2 years ago

5 0

Step-by-step explanation:

I have attached a photo. That might help you out. If still anything is unclear, feel free to let me know in the comments section.

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What is 3x-8-8x=42 answer nowwwwwwwwwwwwwwwwwwwwwwwwwww

eimsori [14]

The answer is x = -10

Explanation:

3x-8-8x=42

3x-8x = 42-8

-5x = 50 ---> do

x = -10

4 0

3 years ago

Graph the equation y=-1(x-3)^2+4

Elena L [17]

Answer:

Direction : Open Down

Vertex : (3,4)

Focus : (3,15/4)

Axis of Symmetry : x=3

Directrix : y=17/4

x : 1,2,3,4,5

y : 0,3,4,3,0

Step-by-step explanation: Graph the parabola using the direction, vertex, focus, and axis symmetry. Graph is included down below!

Hope this helps you out! ☺

3 0

3 years ago

What’s the rule for the rotation given

telo118 [61]

Answer:

A

Step-by-step explanation:

the a and b i put on my triangles should be E and E'. sorry

4 0

3 years ago

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

Two lines, A and B, are represented by the following equations: Line A: y = x − 1 Line B: y = −3x + 11 Which of the following op

OleMash [197]

1) The last one is correct.

6 0

3 years ago

Read 2 more answers