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2 years ago

Multiply and simplify. (x - 1)(x - 1)(x - 1)

1 answer:

3 0

(sorry if it's too late and you've already figured it out, but here you go anyway)
The easiest way to do this is to start by FOILing then add.
So just start with (x-1)(x-1)
(x-1)(x-1)
Front: (x*x) = x^2
Outer: (x*-1) = -x
inner: (-1*x) = -x
Last: (-1*-1) = 1
Added: x^2 -2x +1
Now take that answer and do the same thing with (x-1). It's basically the same thing, just with an added thing you need to multiply.
(x-1)(x^2-2x+1)
(x*x^2) = x^3
(x*2x) = 2x^2
(x*1) = x
(-1*x^2) = -x^2
(-1*-2x) = 2x
(-1*1) = -1
Now add everything together:
x^3+2x^2+x-x^2+2x-1
The answer is:
x^3+x^2+3x-1

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Luba_88 [7]

The answer is 1/7 :)

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2 years ago

Read 2 more answers

HELP!! Algebra help!! Will give stars thank u so much <333

Anna35 [415]

Answers:

Part a) $\bf{\sqrt{x^2+(x^2-3)^2}$
Part b) 3
Part c) 2.24
Part d) 1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0

2 years ago

The quotient of a number w and(-8), which is then increased by 7

Colt1911 [192]

Okay, since quotient is the answer to a division problem, you will divide w and -8. Increased means to add so, you will +7.

w/-8 +7 is what the expression looks like.

5 0

3 years ago

The sum of the first n terms of an arithmetic sequence is n/2(4n + 20).

lesantik [10]

Answer:

(a).

$S_{n} = \frac{n}{2} (4n + 20) \\ \\S _{n - 1} = \frac{(n - 1)}{2} (4n - 4 + 20) \\ \\ S _{n - 1} = \frac{(n - 1)}{2} (4n + 16) \\ \\S _{n - 1} = \frac{(n - 1)(4n + 16)}{2} \\ \\ { \boxed{S _{n - 1} = {2 {n}^{2} + 6n - 8}}} \\$

(b).

from general equation:

$S _{n - 1} = \frac{(n - 1)}{2} (4n + 16)$

first term is 4n

common difference:

$16 = \{(n - 1) - 1 \}d \\ 16 = (n - 2)d \\ d = \frac{16}{n - 2}$

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3 years ago

4x-9y=9

Alexxandr [17]

Multiply both sides of the second equation by 4. That will give you -4x in the second equation which when added to 4x of the first equation will eliminate x.

Second equation:
-x + 3y = 6

Multiply the second equation by 4 on both sides:
-4x + 12y = 26

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3 years ago