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4 years ago

How is my graphs line be for this question?

Mathematics

1 answer:

tekilochka [14]4 years ago

3 0

Looking at the table everything is correct if that is what you are asking.

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Solve the following inequality -6 > - 3c

Inessa05 [86]

Dividing both sides by-3
the division sign will still be the same
your answer is c>2

4 0

4 years ago

Is -6.333 greater less equal to -6.375

Leviafan [203]

6.375 is greater than 6.333

3 0

3 years ago

Read 2 more answers

What value of x satisfies the equation 5(x – 3) – 2(x + 1) = 4?

hjlf

Answer:

x = 7

Step-by-step explanation:

To solve:

Distribute 5 among everything inside the first parentheses. You'll get 5x - 15.

Next distribute -2 among everything in the second parentheses. You'll get

-2x - 2.

All together: 5x - 15 -2x - 2 = 4

Now combine like terms:

5x - 2x = 3x and -15 - 2 = - 17

All together: 3x - 17 = 4

Add 17 on both sides to get 21, and then divide both sides by 3. Answer is x = 7.

________________________________

Last, don't forget to check your work.

You can plug in 7 for x to get:

5(7 - 3) = 20

5 × 7 = 35

5 × -3 = -15

35 - 15 = 20

-2(7 + 1) = -16

-2 × 7 = -14

-2 × 1 = - 2

-14 -2 = -16

20 - 16 = 4

⬆⬆⬆Therefore this is correct.

Sorry this is a bit lengthy, but hope this helps :)

5 0

3 years ago

If the #2 pencil is the most popular, why’s it still #2?

JulsSmile [24]

Never thought of that

3 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