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2 years ago

Factorise 24e^2-28e-12

Mathematics

1 answer:

Helga [31]2 years ago

8 0

Answer:

4(2e - 3)(3e + 1)

Step-by-step explanation:

Given

24e² - 28e - 12 ← factor out 4 from each term

= 4(6e² - 7e - 3) ← factor the quadratic

Consider the factors of the product of the e² term and the constant term which sum to give the coefficient of the e- term.

product = 6 × - 3 = - 18 and sum = - 7

The factors are - 9 and + 2

Use these factors to split the e- term

6e² - 9e + 2e - 3 ( factor the first/second and third/fourth terms )

= 3e(2e - 3) + 1 (2e - 3) ← factor out (2e - 3) from each term

= (2e - 3)(3e + 1)

Then

24e² - 28e - 12 = 4(2e - 3)(3e + 1) ← in factored form

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3x+2y=11 y=5x-1 SUBSTITUTION WORTH 10 POINTS

cupoosta [38]

X=1 and y=4. first I did
3x+2y=11
y=5x-1
and 3x+2(5x-1)=11
=3x+10x-2=11
=13x-2=11
cancel 2 out
13x=13
x=1. then simply add x in the original y equation,
y=5x-1
y=5(1)-1
y=5-1
y=4.
check your work
3(1)+2(4)=11
4=5(1)-1
again X=1 AND Y=4
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IRISSAK [1]

Answer:

x = 10

Step-by-step explanation:

In the last step, you can see that the fraction $\frac{1}{5}$ has been multiplied by its reciprocal $\frac{5}{1}$ , making it cancel out. The reciprocal has been multiplied to both sides, so all you need to do is multiply $\frac{6}{3}$ · $\frac{5}{1}$ :

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