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Verizon [17]
3 years ago
5

Please Please Please Help Me

Mathematics
1 answer:
maks197457 [2]3 years ago
4 0

Answer:

  1. (x-y^2)(x^2 +xy^2 +y^4)
  2. (a^2 +b)(a^4 -a^2b +b^2)
  3. (m^3-n)(m^6 +m^3n +n^2)
  4. (p+k^3)(p^2 -pk^3 +k^6)
  5. (a^2+b^3)(a^4 -a^2b^3 +b^6)
  6. (x-y)(x^2 +xy +y^2)(x^6 +x^3y^3 +y^6)

Step-by-step explanation:

In every case, the factorization makes use of the standard form for factoring the sum or difference of cubes:

  • a^3 +b^3 = (a +b)(a^2 -ab +b^2)
  • a^3 -b^3 = (a -b)(a^2 +ab +b^2)

1. a=x, b=y^2. Use the formula for the difference.

2. a^2 ⇒ a, b = b. Use the formula for the sum.

3. a=m^3, b=n. Use the formula for the difference.

4. a=p b=k^3. Use the formula for the sum.

5. a^2 ⇒ a, b^3 ⇒ b. Use the formula for the sum.

6. a=x^3, b=y^3. Use the formula for the difference. When you do, the first factor is the difference x^3 -y^3, which can be factored using the difference formula again with a=x, b=y.

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PLEASE HELP ILL MARK BRAINLIEST !!!
Alexus [3.1K]

Answer:

a) the common difference is 20

b) x_8=115 , x_{12}=195

c) the common difference is -13

d) a_{12}=52, a_{15}=13

Step-by-step explanation:

a) what is the common difference of the sequence xn

Looking at the table, we get x_3=16, x_4=36 and x_5= 56

Deterring the common difference by subtracting x_4 from x_3 we get

36-16 =20

So, the common difference is 20

b) what is x_8? what is x_12

The formula used is: x_n=x_1+(n-1)d

We know common difference d= 20, we need to find x_1

Using x_3=16 we can find x_1

x_n=x_1+(n-1)d\\x_3=x_1+(3-1)d\\15=x_1+2(20)\\15=x_1+40\\x_1=15-40\\x_1=-25

So, We have x_1 = -25

Now finding x_8

x_n=x_1+(n-1)d\\x_8=x_1+(8-1)d\\x_8=-25+7(20)\\x_8=-25+140\\x_8=115

So, \mathbf{x_8=115}

Now finding x_{12}

x_n=x_1+(n-1)d\\x_{12}=x_1+(12-1)d\\x_{12}=-25+11(20)\\x_{12}=-25+220\\x_{12}=195

So, \mathbf{x_{12}=195}

c) what is the common difference of the sequence a_m

Looking at the table, we get a_7=104, a_8=91 and a_9= 78

Deterring the common difference by subtracting a_7 from a_8 we get

91-104 =-13

So, the common difference is -13

d) what is a_12? what is a_15?

The formula used is: a_n=a_1+(n-1)d

We know common difference d= -13, we need to find a_1

Using a_7=104 we can find x_1

a_n=a_1+(n-1)d\\a_7=a_1+(7-1)d\\104=a_1+7(-13)\\104=a_1-91\\a_1=104+91\\a_1=195

So, We have a_1 = 195

Now finding a_{12} , put n=12

a_n=a_1+(n-1)d\\a_{12}=a_1+(12-1)d\\a_{12}=195+11(-13)\\a_{12}=195-143\\a_{12}=52

So, \mathbf{a_{12}=52}

Now finding a_{15} , put n=15

a_n=a_1+(n-1)d\\a_{15}=a_1+(15-1)d\\a_{15}=195+14(-13)\\a_{15}=195-182\\a_{15}=13

So, \mathbf{a_{15}=13}

5 0
3 years ago
Given: 3 − 2(x + 4) = 13 prove: x = −9
Brums [2.3K]

Answer:

I hope this helps:))

Step-by-step explanation:

4 0
3 years ago
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Evaluate each expression<br><br> g5-h3 if g=2 and h=7
vovangra [49]
Look at it this way:
g=2
H=7
so it would be 2*5-7*3
than just do the multiplication first so 2*5=10 and 7*3=21 than subtract the two 10-21=-11
7 0
3 years ago
Please help me outtttt
QveST [7]

Answer:

C

Step-by-step explanation:

6 0
3 years ago
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1 + tanx / 1 + cotx =2
Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

1 + tan(x) + tan^2(x) = 2 tan(x)

Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

(tan(x) - 1/2)^2 = -3/4

Take the square root of both sides:

tan(x) - 1/2 = (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

tan(x) = 1/2 + (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer:  x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

4 0
3 years ago
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