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jekas [21]
3 years ago
5

What is the sum of this series 20En-1 (5-n)-220-110-95110

Mathematics
2 answers:
mihalych1998 [28]3 years ago
5 0

The person in the comments is correct 100%

RUDIKE [14]3 years ago
3 0
I'm doing that practice rn too tell me if you got the answer!

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Maya has 120 caramel apples to sell. Each caramel apple is covered with one toppin.
EleoNora [17]

they got sold by her

Step-by-step explanation:

lol she sold her apples to people

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Alicia has invented a new app that two companies are interested in purchasing for a 2-year contract.
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(x+2/x-7) - (x^2+4x+13/x^2-4x-21)
olya-2409 [2.1K]

Answer:

x = -2.98079 or x = -1.15272 or x = 0.892002 or x = 4.24151

Step-by-step explanation:

Solve for x:

-x^2 + x + 14 + 2/x - 13/x^2 = 0

Bring -x^2 + x + 14 + 2/x - 13/x^2 together using the common denominator x^2:

(-x^4 + x^3 + 14 x^2 + 2 x - 13)/x^2 = 0

Multiply both sides by x^2:

-x^4 + x^3 + 14 x^2 + 2 x - 13 = 0

Multiply both sides by -1:

x^4 - x^3 - 14 x^2 - 2 x + 13 = 0

Eliminate the cubic term by substituting y = x - 1/4:

13 - 2 (y + 1/4) - 14 (y + 1/4)^2 - (y + 1/4)^3 + (y + 1/4)^4 = 0

Expand out terms of the left hand side:

y^4 - (115 y^2)/8 - (73 y)/8 + 2973/256 = 0

Add (sqrt(2973) y^2)/8 + (115 y^2)/8 + (73 y)/8 to both sides:

y^4 + (sqrt(2973) y^2)/8 + 2973/256 = (sqrt(2973) y^2)/8 + (115 y^2)/8 + (73 y)/8

y^4 + (sqrt(2973) y^2)/8 + 2973/256 = (y^2 + sqrt(2973)/16)^2:

(y^2 + sqrt(2973)/16)^2 = (sqrt(2973) y^2)/8 + (115 y^2)/8 + (73 y)/8

Add 2 (y^2 + sqrt(2973)/16) λ + λ^2 to both sides:

(y^2 + sqrt(2973)/16)^2 + 2 λ (y^2 + sqrt(2973)/16) + λ^2 = (73 y)/8 + (sqrt(2973) y^2)/8 + (115 y^2)/8 + 2 λ (y^2 + sqrt(2973)/16) + λ^2

(y^2 + sqrt(2973)/16)^2 + 2 λ (y^2 + sqrt(2973)/16) + λ^2 = (y^2 + sqrt(2973)/16 + λ)^2:

(y^2 + sqrt(2973)/16 + λ)^2 = (73 y)/8 + (sqrt(2973) y^2)/8 + (115 y^2)/8 + 2 λ (y^2 + sqrt(2973)/16) + λ^2

(73 y)/8 + (sqrt(2973) y^2)/8 + (115 y^2)/8 + 2 λ (y^2 + sqrt(2973)/16) + λ^2 = (2 λ + 115/8 + sqrt(2973)/8) y^2 + (73 y)/8 + (sqrt(2973) λ)/8 + λ^2:

(y^2 + sqrt(2973)/16 + λ)^2 = y^2 (2 λ + 115/8 + sqrt(2973)/8) + (73 y)/8 + (sqrt(2973) λ)/8 + λ^2

Complete the square on the right hand side:

(y^2 + sqrt(2973)/16 + λ)^2 = (y sqrt(2 λ + 115/8 + sqrt(2973)/8) + 73/(16 sqrt(2 λ + 115/8 + sqrt(2973)/8)))^2 + (4 (2 λ + 115/8 + sqrt(2973)/8) (λ^2 + (sqrt(2973) λ)/8) - 5329/64)/(4 (2 λ + 115/8 + sqrt(2973)/8))

To express the right hand side as a square, find a value of λ such that the last term is 0.

This means 4 (2 λ + 115/8 + sqrt(2973)/8) (λ^2 + (sqrt(2973) λ)/8) - 5329/64 = 1/64 (512 λ^3 + 96 sqrt(2973) λ^2 + 3680 λ^2 + 460 sqrt(2973) λ + 11892 λ - 5329) = 0.

Thus the root λ = 1/48 (-3 sqrt(2973) - 115) + 1/12 (-i sqrt(3) + 1) ((3 i sqrt(10705335) - 8327)/2)^(1/3) + (173 (i sqrt(3) + 1))/(3 2^(2/3) (3 i sqrt(10705335) - 8327)^(1/3)) allows the right hand side to be expressed as a square.

(This value will be substituted later):

(y^2 + sqrt(2973)/16 + λ)^2 = (y sqrt(2 λ + 115/8 + sqrt(2973)/8) + 73/(16 sqrt(2 λ + 115/8 + sqrt(2973)/8)))^2

Take the square root of both sides:

y^2 + sqrt(2973)/16 + λ = y sqrt(2 λ + 115/8 + sqrt(2973)/8) + 73/(16 sqrt(2 λ + 115/8 + sqrt(2973)/8)) or y^2 + sqrt(2973)/16 + λ = -y sqrt(2 λ + 115/8 + sqrt(2973)/8) - 73/(16 sqrt(2 λ + 115/8 + sqrt(2973)/8))

Solve using the quadratic formula:

y = 1/8 (sqrt(2) sqrt(16 λ + 115 + sqrt(2973)) + sqrt(2) sqrt((10252 - 32 sqrt(2973) λ - 256 λ^2 + 292 sqrt(2) sqrt(16 λ + 115 + sqrt(2973)))/(16 λ + 115 + sqrt(2973)))) or y = 1/8 (sqrt(2) sqrt(16 λ + 115 + sqrt(2973)) - sqrt(2) sqrt((10252 - 32 sqrt(2973) λ - 256 λ^2 + 292 sqrt(2) sqrt(16 λ + 115 + sqrt(2973)))/(16 λ + 115 + sqrt(2973)))) or y = 1/8 (sqrt(2) sqrt((10252 - 32 sqrt(2973) λ - 256 λ^2 - 292 sqrt(2) sqrt(16 λ + 115 + sqrt(2973)))/(16 λ + 115 + sqrt(2973))) - sqrt(2) sqrt(16 λ + 115 + sqrt(2973))) or y = 1/8 (-sqrt(2) sqrt(16 λ + 115 + sqrt(2973)) - sqrt(2) sqrt((10252 - 32 sqrt(2973) λ - 256 λ^2 - 292 sqrt(2) sqrt(16 λ + 115 + sqrt(2973)))/(16 λ + 115 + sqrt(2973)))) where λ = 1/48 (-3 sqrt(2973) - 115) + 1/12 (-i sqrt(3) + 1) ((3 i sqrt(10705335) - 8327)/2)^(1/3) + (173 (i sqrt(3) + 1))/(3 2^(2/3) (3 i sqrt(10705335) - 8327)^(1/3))

Substitute λ = 1/48 (-3 sqrt(2973) - 115) + 1/12 (-i sqrt(3) + 1) ((3 i sqrt(10705335) - 8327)/2)^(1/3) + (173 (i sqrt(3) + 1))/(3 2^(2/3) (3 i sqrt(10705335) - 8327)^(1/3)) and approximate:

y = -3.23079 or y = -1.40272 or y = 0.642002 or y = 3.99151

Substitute back for y = x - 1/4:

x - 1/4 = -3.23079 or y = -1.40272 or y = 0.642002 or y = 3.99151

Add 1/4 to both sides:

x = -2.98079 or y = -1.40272 or y = 0.642002 or y = 3.99151

Substitute back for y = x - 1/4:

x = -2.98079 or x - 1/4 = -1.40272 or y = 0.642002 or y = 3.99151

Add 1/4 to both sides:

x = -2.98079 or x = -1.15272 or y = 0.642002 or y = 3.99151

Substitute back for y = x - 1/4:

x = -2.98079 or x = -1.15272 or x - 1/4 = 0.642002 or y = 3.99151

Add 1/4 to both sides:

x = -2.98079 or x = -1.15272 or x = 0.892002 or y = 3.99151

Substitute back for y = x - 1/4:

x = -2.98079 or x = -1.15272 or x = 0.892002 or x - 1/4 = 3.99151

Add 1/4 to both sides:

Answer: x = -2.98079 or x = -1.15272 or x = 0.892002 or x = 4.24151

7 0
3 years ago
Read 2 more answers
Joaquim is baking giant cookies for the school bake sale. They will be sold for $20 for one large cookie or $20 for three small
velikii [3]
Small i believe
Because sure the large is bigger but you get more with the small cookies
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3 years ago
2. Use the distributive property to rewrite the following expression.<br> 3(x+5)
baherus [9]

The answer is 3x+15

You multiply the x and 5 by 3 to get 3x and 15

6 0
3 years ago
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