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3 years ago

Find all the real zeros of f(x)=3x^3-9x^2+3x-9 Can someone help me please

Mathematics

1 answer:

vaieri [72.5K]3 years ago

4 0

Answer:

x=0,3

Step-by-step explanation:

start by dividing every term by 3

x^3-3x^2+x-3

group into 2 terms

(x^3-3x^2)+(x-3)

simplify as much as you can

x^2(x-3)+(x-3)

combine terms

(x^2)(x-3)

x=3, 0

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-3(6 - 3k)<br>I need to Simplify the expression

sergij07 [2.7K]

Answer:

-18+9k

Step-by-step explanation:

If you use the distributive property, you get:

(-3*6) and (-3*-3k)

If you simplify each of them you get:

-18+9k

8 0

3 years ago

Read 2 more answers

Show Your Work<br> Please help I’ll mark brainliest

fgiga [73]

Answer:

(c)

Step-by-step explanation:

3+2+4 = 9, 4+2= 6- 3= 3

3 0

3 years ago

Which is the value of this expression when m=3 and n=-5? (6m^-1 n^0)^-3

iragen [17]

Answer:

The answer is C.

Step-by-step explanation:

I would simplify the expression first.

Equation: (6m^-1)^-3

You can get rid of n^0 because that equals 1.

Any expression raised to the power of -1 equals its reciprocal.

Equation: (6/m)^-3

Equation: (m/6)^3

Final Equation: m^3/216

Now, plug in 3.

(3)^3/216.

27/216 = 1/8

Hope this helps!

6 0

3 years ago

Can anyone help me find the coordinates (?,?) given that the shape is a parallelogram? (15 points)

xxTIMURxx [149]

The missing coordinates of the parallelogram is (m + h, n).

Solution:

Diagonals of the parallelogram bisect each other.

Solve using mid-point formula:

$$\text{Midpoint} =\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right )$

Here $x_1=m, y_1=n, x_2=h, y_2=0$

$$=\left( \frac{m+h}{2}, \frac{n+0}{2}\right )$

$$\text{Midpoint} =\left( \frac{m+h}{2}, \frac{n}{2}\right )$

<u>To find the missing coordinate:</u>

Let the missing coordinates by x and y.

Here $x_1=0, y_1=0, x_2=x, y_2=y$

$$\text{Midpoint}=\left( \frac{0+x}{2}, \frac{0+y}{2}\right )$

$$\left( \frac{m+h}{2}, \frac{n}{2}\right )=\left( \frac{0+x}{2}, \frac{0+y}{2}\right )$

$$\left( \frac{m+h}{2}, \frac{n}{2}\right )=\left( \frac{x}{2}, \frac{y}{2}\right )$

Now equate the x-coordinate.

$$ \frac{m+h}{2}=\frac{x}{2}$

Multiply by 2 on both sides of the equation, we get

m + h = x

x = m + h

Now equate the y-coordinate.

$$\frac{n}{2} = \frac{y}{2}$

Multiply by 2 on both sides of the equation, we get

n = y

y = n

Hence the missing coordinates of the parallelogram is (m + h, n).

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3 years ago

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