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3 years ago

I really need help with this equation as soon as possible :D

1 answer:

7 0

The subtraction sign in front of the second set of parenthesis changes the signs on the terms inside those parenthesis. So this could be rewritten as $x-3x^4+2x^2+x^3-4x^6$ . None of the exponents on the x are the same, so we will just express it in terms of descending powers of x. $-4x^6-3x^4+x^3+2x^2+x$ . This is also the same as $4x^6+3x^4-x^3-2x^2-x$ if you don't like to start with a negative. Either one is correct.

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Help plz plz plz plz .

nikklg [1K]

The y intercept is where the line hits the verticals line of the graph. Hope this helps!

7 0

3 years ago

How do I do this when I first learn it

oksano4ka [1.4K]

You would put in 12 for x since there are 12 months in a year. Then you would multiply 12 by 35, and add 70.

3 0

3 years ago

Read 2 more answers

PLEASE ANSWER THIS AS SOON AS POSSIBLE

monitta

Answer:

-3

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

13.

Artist 52 [7]

Answer:

5 units

Step-by-step explanation:

An isosceles triangle is a triangle with two legs that have the same length. The perimeter of a triangle is the sum of the lengths of all sides of the triangle. Now taking this into account, we know that:

2L + B = 14 units

Where:

L is the measure of one leg

B is the measure of the base

Since two legs are the same and the base is 1 less, this means the measure of each leg would be:

B = L -1

Now we have two equations:

2L + B = 14 units

B = L- 1

We plug one equation into the other and make 1 equation:

2L + (L-1) = 14 units

Get rid of the parentheses:

2L + L - 1 = 14

Combine like terms:

3L - 1 = 14

Add 1 to both sides of the equation:

3L - 1 + 1 = 14 + 1

3L = 15

Divide both sides by 3:

3L/3 = 15/3

L = 5

So the length of a leg is 5 units

Let's check!

B = L - 1

B = 5 - 1

B = 4

Then we use that to solve for the perimeter:

2L + B

2(5) + 4

10 + 4 = 14

7 0

3 years ago

Find the volumes of the solids generated by revolving the triangle with vertices (2, 2), (2, 6), and (5, 6) about a) the x-a

Vesna [10]

About the $x$ -axis (washer method):

$\displaystyle\pi\int_2^5\left(6^2-\left(\frac43x-\frac23\right)^2\right)\,\mathrm dx=\frac{16\pi}9\int_2^5(20+x-x^2)\,\mathrm dx=\boxed{56\pi}$

About the $y$ -axis (shell method):

$\displaystyle2\pi\int_2^5x\left(6-\left(\frac43x-\frac23\right)\right)\,\mathrm dx=\frac{8\pi}3\int_2^5x(5-x)\,\mathrm dx=\boxed{36\pi}$

About $x=7$ (shell method):

$\displaystyle2\pi\int_2^5(7-x)\left(6-\left(\frac43x-\frac23\right)\right)\,\mathrm dx=\frac{8\pi}3\int_2^5(35-12x+x^2)\,\mathrm dx=\boxed{48\pi}$

About $y=2$ (washer method):

$\displaystyle\pi\int_2^5\left((6-2)^2-\left(\frac43x-\frac23-2\right)^2\right)\,\mathrm dx=\frac{16\pi}9\int_2^5(5+4x-x^2)\,\mathrm dx=\boxed{32\pi}$

7 0

3 years ago