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1 year ago

Simplify the following expressions

Mathematics

2 answers:

Allushta [10]1 year ago

7 0

Answer:

See below.

Step-by-Step Explanation:

5) 8x+9-6x-7

8x + 2 - 6x (Subtract the numbers)

2x + 2 (Combine like terms)

2 (x + 1) (Common factor - Factor by grouping twice)

6) 2y-6+5y+9

2y + 3 + 5y (Add the numbers)

7y + 3 (Combine like terms)

7) 3x+8+4x-9

3x - 1 + 4x (Subtract the numbers)

7x - 1 (Combine like terms)

8) 9y+16-12y+24

9y + 40 - 12y (Add the numbers)

-3y + 40 (Combine like terms)

9) 7g-5-8g+12

7g + 7 - 8g (Add the numbers)

-g + 7 (Combine like terms)

10) 11c-8+5c-8

11c - 16 + 5c (Subtract the numbers)

16c - 16 (Combine like terms)

16 (c - 1) (Common factor - Factor by grouping twice)

11) 5(2x + 3) (This one was a bit blurry, I am assuming that's a addition sign.)

10x + 15 (Distribute)

12) 5(-2x+3)

-10x + 15 (Distribute)

Hope this helps!

Effectus [21]1 year ago

4 0

Answer: those are the answers its a long time.

Question 5 Answer: 2(x+1)

8x+9−6x−7

Multiply and combine like terms.

2x+2

Factor out 2.

2(x+1)

Questio 6 Answer : 7y + 3

2y−6+5y+9

Combine 2y and 5y to get 7y.

7y−6+9

Add −6 and 9 to get 3.

7y+3

Question 7 Answer: 3x+8+4x−9

Combine 3x and 4x to get 7x.

7x+8−9

Subtract 9 from 8 to get −1.

7x−1

Question 8 Answer: - 3y + 40

9y+16−12y+24

Combine 9y and −12y to get −3y.

−3y+16+24

Add 16 and 24 to get 40.

−3y+40

Question 9 Answer: -g + 7

7g−5−8g+12

Combine 7g and −8g to get −g.

−g−5+12

Add −5 and 12 to get 7.

−g + 7

Question 10 Answer: 16 ( c - 1 )

11c−8+5c−8

Combine 11c and 5c to get 16c.

16c−8−8

Subtract 8 from −8 to get −16.

16c - 16

Question 11 Answer: 10x +15

5(2x+3)

Use the distributive property to multiply 5 by 2x+3.

10x+15

Question 12 Answer: 15 - 10x

5(−2x+3)

Use the distributive property to multiply 5 by −2x+3.

−10x+15

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Follows are the solution to the given points:

Step-by-step explanation:

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$u = x \\\\dv = e^{(t-1)x}\\\\dx =dx \\\\v= \frac{e^{(t-1)x}}{t-1}\\\\M(t) =[\frac{e^{(t-1)x}}{t-1}]^{-\infty}_{0} -\int^{\infty}_{0} \frac{e^{(t-1)x}}{t-1} \ dx\\\\$

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$E(X^n)=?$

Using formula: $E(X^n)= M^{n}_{X}(0)$

form point (a):

$\to M_{X}(t)=\frac{1}{(t-1)^2}$

Differentiating the value with respect of t

$M'_{X}(t)=\frac{-2}{(t-1)^3}$

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$M'_{X}(0)=\frac{-2}{(0-1)^3}= \frac{-2}{(-1)^3}= \frac{-2}{-1}=2\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M'''_{X}(t)=\frac{(-2)(-3)(-4)}{(t-1)^5}\\\\M''_{X}(0)=\frac{(-2)(-3)(-4)}{(0-1)^5}= \frac{24}{(-1)^5}= \frac{24}{-1}=-24\\\\\therefore \\\\M^{K}_{X} (t)=\frac{(-2)(-3)(-4).....(k+1)}{(t-1)^{k+2}}\\\\$

$M^{K}_{X} (0)=\frac{(-2)(-3)(-4).....(k+1)}{(-1)^{k+2}}\\\\\therefore\\\\E(X^n) = \frac{(-2)(-3)(-4).....(n+1)}{(-1)^{n+2}}\\\\$

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