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anyanavicka [17]
3 years ago
8

What is the graph of y > x2-9.​

Mathematics
1 answer:
Yakvenalex [24]3 years ago
8 0

Answer:

The 1st graph

Step-by-step explanation:

When you graph the equation in a graphing calc, you should be able to see that the graph most similar would be the 1st one.

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Janelle was trying to find the distance between (3,7) and (9,6) in the coordinate plane. She knew the formula was D=√(9 - 3)^2 +
s2008m [1.1K]
Hello : 
incorrect  answer because : 
\sqrt{ a^{2} + b^{2} }  \neq   \sqrt{ a^{2} } + \sqrt{ b^{2}}  \neq a+b
but : 
\sqrt{ (a+b)^{2} } = a+b........a+b <span>≥ 0</span>
5 0
3 years ago
Consider the expression √(√625).
const2013 [10]

Answer:

5

Step-by-step explanation:

Square Root - Finding a number that multiplies itself twice into the number within the square root

With this meaning, we need to find a number that multiplies itself into 625.

Finding calculations using exponents will help,

25^2 is equal to 625. Therefore \sqrt{25}

We are not finished, as there is another square root right after, empowering the parenthesis, so what number multiplies itself equals 25?

This would be 5, therefore \sqrt{(\sqrt{625}) } is equal to 5.

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
3 years ago
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30%

Step-by-step explanation:

5+2=7+3=10

3/10=30%

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Which is the dependent variable in 4x^2-5/6x-9=y if y=f(x)
olga55 [171]

Answer:

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Step-by-step explanation:

The expression

  y = f(x)

tells you that y is the dependent variable, and that it depends on x, the independent variable. The <em>independent</em> variable is always the function argument. Any variable that depends on that is the <em>dependent</em> variable.

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