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2 years ago

7

Which of the following equations would have 2 x-intercepts

1 answer:

Radda [10]2 years ago

5 0

Answer:

None of these

Step-by-step explanation:

because if we plug in every time we will gwt one solution of X

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Makovka662 [10]

Let $u=x^2-4$ and $v=4x-5$ . By the product rule,

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=\dfrac{\mathrm d(u^5)}{\mathrm dx}v^4+u^5\dfrac{\mathrm d(v^4)}{\mathrm dx}$

By the power rule, we have $(u^5)'=5u^4$ and $(v^4)'=4v^3$ , but $u,v$ are functions of $x$ , so we also need to apply the chain rule:

$\dfrac{\mathrm d(u^5)}{\mathrm dx}=5u^4\dfrac{\mathrm du}{\mathrm dx}$

$\dfrac{\mathrm d(v^4)}{\mathrm dx}=4v^3\dfrac{\mathrm dv}{\mathrm dx}$

and we have

$\dfrac{\mathrm du}{\mathrm dx}=2x$

$\dfrac{\mathrm dv}{\mathrm dx}=4$

So we end up with

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=10xu^4v^4+16u^5v^3$

Replace $u,v$ to get everything in terms of $x$ :

$\dfrac{\mathrm d((x^2-4)^5(4x-5)^4)}{\mathrm dx}=10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3$

We can simplify this by factoring:

$10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3=2(x^2-4)^4(4x-5)^3\bigg(5x(4x-5)+8(x^2-4)\bigg)$

$=2(x^2-4)^4(4x-5)^3(28x^2-57)$

7 0

3 years ago

Please help I will vote Brainiest hurry!! Ignore that they r already answered I can't get rid of them and they r probably wrong

GaryK [48]

Answer:

i am prtty sure it gores dot 3, dot 2, and dot 4

Step-by-step explanation:

dont quote me on that tho

5 0

3 years ago

The average number of acres burned by all wildfires in the united states is 780 acres with standard deviation 500 acres. of cour

shusha [124]

Mmmmmmmmmmmmmmmmmmmmmmmmmmmm

7 0

3 years ago

What is the meaning of backtracking?

tia_tia [17]

Answer:

Retracing your steps

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

What is the distance from Cto D?

IRINA_888 [86]

Given:

The two points C(2,-1) and D(5,3) on a coordinate plane.

To find:

The distance from C to D.

Solution:

The distance between two points is defined by the distance formula.

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The two points are C(2,-1) and D(5,3). Using the distance formula, we get

$CD=\sqrt{(5-2)^2+(3-(-1))^2}$

$CD=\sqrt{(3)^2+(3+1)^2}$

$CD=\sqrt{(3)^2+4^2}$

On further simplification, we get

$CD=\sqrt{9+16}$

$CD=\sqrt{25}$

$CD=5$

The distance from point C to D is 5 units. Therefore, the correct option is A.

8 0

2 years ago