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Anna [14]
2 years ago
6

Which of the following are considered assets?

Mathematics
2 answers:
nordsb [41]2 years ago
8 0

Answer:

Number 1, i think

Step-by-step explanation:

algol132 years ago
7 0

Answer:

III and II because those are MANDATORY everday. Assets.

Step-by-step explanation:

Assests mean what is mandatory in everyday life.

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Is y = |x| a function
Doss [256]

Answer:

Yes

Step-by-step explanation:

A function is a relation for which each value from the set of the first components of the ordered pairs is associated with exactly one value from the set of second components of the ordered pair.

Hope this helps!

8 0
3 years ago
Read 2 more answers
No clue what I'm doing help
Umnica [9.8K]
-32=y+4 is the answer
6 0
3 years ago
(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
2 years ago
Read 2 more answers
A penguin walks 15 feet in 9 seconds.
shtirl [24]

Answer:

12 seconds.

Step-by-step explanation:

If you put it like 15/9 = 20/x, then cross multiply and divide you do 9 times 20 you get 180, then you divide 180 by 15 and you get 12. Hope this helps!

4 0
3 years ago
Read 2 more answers
Which expression is equivalent to one over five m − 20? (4 points)
exis [7]

Answer:

The equivalent to one over five m − 20 is one over five (m − 100) ⇒ B

Step-by-step explanation:

Let us solve the question

∵ One over five means \frac{1}{5}

∴ One over five m - 20 = \frac{1}{5} m - 20

→ By using the distributive property, take one over five as a common factor

  from both terms

∴ \frac{1}{5} m - 20 = \frac{1}{5} (\frac{\frac{1}{5}m}{\frac{1}{5}} - \frac{20}{\frac{1}{5}})

→ Simplify the bracket

∵  \frac{1}{5} m ÷ \frac{1}{5} = \frac{1}{5} m × 5 = m

∵ 20 ÷  \frac{1}{5}  = 20 × 5 = 100

∴  \frac{1}{5} (\frac{\frac{1}{5}m}{\frac{1}{5}} - \frac{20}{\frac{1}{5}}) = \frac{1}{5} (m - 100)

∴ \frac{1}{5} m - 20 = \frac{1}{5} (m - 100)

The equivalent to one over five m − 20 is one over five (m − 100)

4 0
3 years ago
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