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Bad White [126]
3 years ago
13

Factorise :12(x²+7)²-8(x²+7)(2x-1)-15(2x-1)²

Mathematics
2 answers:
Readme [11.4K]3 years ago
7 0
12(x^2+7)2−8(x^2+7)(2x−1)−15(2x−1)^2

Distribute:

=12x^4+168x^2+588+−16x^3+8x^2+−112x+56+−60x^2+60x+−15

Combine Like Terms:

=12x^4+168x^2+588+−16x^3+8x^2+−112x+56+−60x^2+60x+−15

=(12x^4)+(−16x^3)+(168x^2+8x^2+−60x^2)+(−112x+60x)+(588+56+−15)

=12x^4+−16x^3+116x^2+−52x+629

adelina 88 [10]3 years ago
3 0
12(x²+7)²-8(x²+7)(2x-1)-15(2x-1)²

(x²+7)=a
(2x-1)=b

12a²-8ab-15b²=12a²-18ab+10ab-15b²=6a(2a-3b)+5b(2a-3b)=(2a-3b)(6a+5b)

[ 2(x²+7)-3(2x-1) ] [ 6(x²+7)+5(2x-1) ]=
( 2x²+14-6x+3 ) ( 6x²+42+10x-5) )=
(2x²-6x+17)(6x²+10x+37)=
2x²×6x² -  6x × 6x² +17×6x² + 2x²×10x - 6x ×10x+17×10x+2x²×37-6x ×37+17 × 37=
12x⁴ -36x³+102x²+20x³-60x²+170x+74x²-222x+629=
12x⁴ -36x³+20x³+102x²-60x²+74x²+170x-222x+629=
<u>=12x⁴-16x³+116x²-52x+629</u>
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A triangle has side lengths of (1.3f+2.9g)(1.3f+2.9g) centimeters, (1.1f-7.7h)(1.1f−7.7h) centimeters, and (2.3h+1.5g)(2.3h+1.5g
VikaD [51]

Answer:

Perimeter of the triangle = (2.4f+4.4g-5.4h) centimeters.

Step-by-step explanation:

It is given that, a triangle has side lengths of (1.3f+2.9g) cm, (1.1f-7.7h)cm, and (2.3h+1.5g) cm.

We have to find the expression which represents the perimeter, in centimeters, of the triangle.

Perimeter of a triangle = Sum of its all the sides.

Perimeter=(1.3f+2.9g)+(1.1f-7.7h)+(2.3h+1.5g)

On combining like terms, we get

Perimeter=(1.3f+1.1f)+(2.9g+1.5g)+(-7.7h+2.3h)

Perimeter=2.4f+4.4g+(-5.4h)

Perimeter=2.4f+4.4g-5.4h

Therefore, the perimeter of the triangle is (2.4f+4.4g-5.4h) centimeters.

8 0
2 years ago
Find the horizontal and vertical asymptotes of​ f(x). ​f(x) equals = StartFraction 6 x Over x plus 2 EndFraction 6x x+2 Find the
miss Akunina [59]

Answer:

The horizontal asymptote can be described by the line y = 6

The vertical asymptote can be described by the line x = -2

Step-by-step explanation:

* <em>Lets the meaning of vertical and horizontal asymptotes</em>

- <u><em>Vertical asymptotes</em></u> are vertical lines which correspond to the zeroes

  of the denominator of a rational function

- <u><em>A horizontal asymptote</em></u> is a y-value on a graph which a function

 approaches but does not actually reach

- If the degree of the numerator is less than the degree of the

 denominator, then there is a horizontal asymptote at y = 0

- If the degree of the numerator is greater than the degree of the

 denominator, then there is no horizontal asymptote

- If the degree of the numerator is equal the degree of the denominator,

 then there is a horizontal asymptote at y = leading coefficient of the

 numerator ÷ leading coefficient of the denominator

* <em>Lets solve the problem</em>

∵ f(x)=\frac{6x}{x+2}

∵ The numerator is 6x

∵ The denominator is x + 2

∴ The numerator and the denominator have same degree

∵ The leading coefficient of the numerator is 6

∵ The leading coefficient of the denominator is 1

∴ There is a horizontal asymptote at y = 6/1

∴ <em>The horizontal asymptote can be described by the line y = 6</em>

- Put the denominator equal zero to find its zeroes

∵ The denominator is x + 2

∴ x + 2 = 0

- Subtract 2 from both sides

∴ x = -2

∴ <em>The vertical asymptote can be described by the line x = -2</em>

6 0
2 years ago
Read 2 more answers
Hi... I'm a dog <br><br> Factor 64b-16c to identify the equivalent expressions.<br> Choose 2 answers
AleksandrR [38]
8(8b-2c) and 2(32b-8c)
3 0
3 years ago
Determine the value for AB
grin007 [14]

Answer:

x = 2

AB = 14

Step-by-step explanation:

2x(10) = 8(x+3)

20x = 8x + 24

subtract 8x from both sides

12x = 24

x = 2

Check:

4(10) = 8(5)

40 = 40

Correct

2x = 2 x 2 = 4

4 + 10 = 14

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

5 0
2 years ago
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Three boxes have a total weight of 510 pounds. box a weighs twice as much as box
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Work shown above! Box a is 240 lbs Box b is 120 and Box c is 150 lbs hope this helps c:

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