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xz_007 [3.2K]
3 years ago
14

Factor by grouping. 4x^3 − 16x^2 + 12 − 3x

Mathematics
1 answer:
AleksAgata [21]3 years ago
5 0
He first thing that we would really want to consider is that we would definitely want to group this up because it would make things a lot more easier.

(((4 * (x^3)) -  2^4x^2) +  12) -  3x \\ \\ as \ we \ simplify \ this . . . \\ \\  ((2^2x^3 -  2^4x^2) +  12) -  3x \\ \\ Factoring:  \ \boxed{4x^3-16x^2-3x+12 } \\ \\ Factoring: \  4x^2-3 \\ \\

Your answer: \boxed{\bf{  (4x^2 - 3) *(x - 4)}}
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Solve this literary equation:<br><br> 15= -3(2c+d)
adelina 88 [10]

Answer:

15=-6c-3d

Step-by-step explanation:

If 15=-3(2c+d)

=15=(-3×2c) - (-3×d)

=15=-6c-3d

6 0
3 years ago
What is the rate of change from -1 to 4?
lana66690 [7]

Answer:

5

Step-by-step explanation:

-1+(1)= 0

0+(4)=4

1+4=5

5 0
3 years ago
Kelly ran 6.2 miles Monday, 10.95 miles Wednesday, and 8.25 miles on Thursday. How many miles did she run in all?
RUDIKE [14]

Answer:

25.4 miles

Step-by-step explanation:

added them all up

but if i just wants miles than tha answer is 25

3 0
3 years ago
I need help with this geometry question
MArishka [77]

Answer:

Approximately 7.8 units

Step-by-step explanation:

To find the distance between any two points, we can use the distance formula:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

From the graph, we can see that A is (0,0). Let's let this be x₁ and y₁.

B is (-5,6). Let's let this be x₂ and y₂. So, substitute:

d=\sqrt{(-5-0)^2+(6-0)^2}

Simplify:

d=\sqrt{(-5)^2+(6)^2}

Square:

d=\sqrt{25+36}

Add:

d=\sqrt{61}

Take the square root. Use a calculator:

d\approx7.8

So, the distance between them is about 7.8 units.

And we're done!

7 0
3 years ago
URGENT! WILL GIVE BRAINLY FOR CORRECT ANSWER
Minchanka [31]

Answer:

sorry if im wrong i think wrong numbers. plz check

Step-by-step explanation:

The center of dilation of the question is (-4,-3) .

let say that

x0=-4

y0=-3

Label the image A'B'C'

The new coordinate would be

A(-4,-1)

x=4

y=-1

x'=x0+ 2(x - x0)

x'= -4+ 2(-4 +4)

x'=-4

y'=y0+ 2(y - y0)

y'= -3+ 2(-1 +3)

y'=-3 +4= 1

______________________________

B(-4,-3)

x=-4

y=-3

x'=x0+ 2(x - x0)

x'= -4+ 2(-4 +4)

x'=-4

y'=y0+ 2(y - y0)

y'= -3+ 2(-3 +3)

y'=-3

______________________________

C(-1,-3)

x=-1

y=-3

x'=x0+ 2(x - x0)

x'= -4+ 2(-1 +4)

x'=-4 +6= 2

y'=y0+ 2(y - y0)

y'= -3+ 2(-3 +3)

y'=-3

A'(-4,1)

B'(-4,-3)

C'(2,-3)

3 0
2 years ago
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