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Zielflug [23.3K]
3 years ago
15

Adda³b(a²-2a+3) and 2ab(a³+3a²-5b)please answer this question step wise​

Mathematics
1 answer:
Allushta [10]3 years ago
5 0

Step-by-step explanation:

i hope it helps you!! :)))

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-3(x + 4) = (-x - 1)
zaharov [31]

Answer:

x=-5.5

Step-by-step explanation:

-3(x+4)=(-x-1)

1. Distribute

-3x+(-12)=-x-1

2. Simplify

-3x-12=-x-1

-2x-12=-1

-2x=11

x=-11/2

x=-5.5

4 0
3 years ago
Read 2 more answers
A hydro company is laying down power lines in the ground. They have two sections of cable so far that are 325.0 m and 430.0 m lo
Hoochie [10]

The length of extra cable that is required to connect the two pieces of existing cable is equal to 182 meters.

<h3>How to determine the length of extra cable?</h3>

In order to determine the length of extra cable that is required to connect the two pieces of existing cable, we would apply the law of cosine as follows:

B² = A² + C² - 2(A)(C)cosB

Substituting the given parameters into the formula, we have;

B² = 325.0² + 430.0² - 2(325.0)(430.0)cos23

B² = 105,625 + 184,900 - 279,500(0.9205)

B² = 290,525 - 257,279.75

B² = 33,245.25

B = √33,245.25

B = 182.33 ≈ 182 meters.

Read more on cosine law here: brainly.com/question/11000638

#SPJ1

3 0
2 years ago
Point b is between points a and
Kisachek [45]
2x-3+7x-10=32
9x-13=32
9x=32+13
9x=45
x=5
7 0
3 years ago
A. A farmer irrigates her fields with 120,000 cubic yards of water. She has a 1,000-acre farm. How
horsena [70]

Answer:

120 cc yards of water

Step-by-step explanation:

A farmer has 1,000-acre farm. And she irrigates her fields with 120,000 cubic yards of water. We are asked to find how many cubic feet of water does she apply per acre?

1000 acre is irrigated by = 120000 cc yard of water

Hence 1 Acre is irrigated by

= \frac{120000}{1000} cc yards of water

Hence 1 Acre is irrigated by 120 cc yards of water .

8 0
3 years ago
Y=3(x+1)(x-5)<br><br> Explain how to solve these quadratics.
olya-2409 [2.1K]
Y=3(x+1)(x-5)
I'm going to ignore 3 for now, but will add it later.
FOIL the (x+1)(x-5)
F = multiply the First numbers      x and x
O = multiply the Outside numbers      x and -5
I = multiply the Inside number        1 and x
L = multiply the Last numbers        1 and -5
3(x^2 + -5x + 1x + -5)
add -5x and x
3(x^2 + -4x + -5)
multiply the numbers inside the parenthesis by 3
3x^2 + -12x + -15
5 0
3 years ago
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