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3 years ago

12

Can someone help me?

2 answers:

neonofarm [45]3 years ago

8 0

A, B and D are possible

to do that, put -3 as x and 2 as y and check whether the equation is true or not.

good luck

astra-53 [7]3 years ago

4 0

Slope m = 2/3 passing thru (-3, 2)
equation in point slope form
y - y1 = m(x - x1)
so
y - 2 = 2/3(x + 3) -------------answer is A.
y = 2/3 x + 2
y = 2/3 x + 4 --------------answer is D.
or
3y = 2x + 12
2x - 3y = -12 ------------------answer is B

Answers:
A, B and D.

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yuradex [85]

Answer:

Step-by-step explanation:

Polynomial f(x) has the following conditions: zeros of -4 (multiplicity 3), 1 (multiplicity 1), and with f(0) = 320.

The first part zeros of -4 means (x+4) and multiplicity 3 means (x+4)^3.

The second part zeros of 1 means (x-1) and multiplicity 1 means (x-1).

The third part f(0) = 320 means substituting x=0 into (x+4)^3*(x-1)*k =320

(0+4)^3*(0-1)*k = 320

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k = -5

Combining all three conditions, f(x)

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8 0

3 years ago

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podryga [215]

Answer:

$g(x)=-2\sqrt[3]x$

or

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Step-by-step explanation:

Given

$f(x) = \sqrt[3]x$

Required

Write a rule for g(x)

See attachment for grid

From the attachment, we have:

$(x_1,y_1) = (-1,2)$

$(x_2,y_2) = (1,-2)$

We can represent g(x) as:

$g(x) = n * f(x)$

So, we have:

$g(x) = n * \sqrt[3]x$

For:

$(x_1,y_1) = (-1,2)$

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This gives:

$2 = n * -1$

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$n = \frac{2}{-1}$

$n = -2$

To confirm this value of n, we make use of:

$(x_2,y_2) = (1,-2)$

So, we have:

$-2 = n * \sqrt[3]1$

This gives:

$-2 = n * 1$

Solve for n

$n = \frac{-2}{1}$

$n = -2$

Hence:

$g(x) = n * \sqrt[3]x$

$g(x)=-2\sqrt[3]x$

or:

$g(x) = -2f(x)$

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3 years ago

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