All categories

3 years ago

Write the equation for the function of the graph below

Mathematics

1 answer:

riadik2000 [5.3K]3 years ago

6 0

Answer: (1) $y= \sqrt{x+2}-2$ (2) y = (x - 3)³

D: x ≥ -2 [-2, ∞) D: -∞ < x < ∞ (-∞, ∞)

R: y ≥ -2 [-2, ∞) R: -∞ < y < ∞ (-∞, ∞)

Inc: x > 2 (2, ∞) Inc: (-∞, 3) ∪ (3, ∞)

Dec: never Dec: never

Step-by-step explanation:

NOTES:

Domain is the x-values.

Range is the y-values.

From left to right: increasing when going up and decreasing when going down. The anchor and turning point are not considered to be increasing or decreasing.

You might be interested in

How many seats are there in each row?

andrezito [222]

Fives Guys Cheeseburger with Ketchup, Mayo, Grilled Onions, Cheese, and A1 Steak Sauce

7 0

3 years ago

How much money, invested at an interest rate of r % per year compounded continuously, will amount to A dollars after t years? (R

Talja [164]

A=A(1+r/100)^t
a=17000(1+.043)^3
a=19.288.65
a is your money after 3 years

6 0

3 years ago

Enter the value for x that makes the equation 1/2(4x-8)+3x=36 true

MrRissso [65]

Answer:

the correct answer for x is 8

Step-by-step explanation:

since: 1/2 ((4) (8) - 8) + (3) (8) = 36

4 0

3 years ago

Elisa is making candles. she needs 5 ounces of wax for each candle. she has 25 ounces of wax. how many candles can she make

Lena [83]

Five candles. 25 ounces divided by 5 ounces equals 5 candles.

8 0

4 years ago

Read 2 more answers

Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}

miskamm [114]

Answer:

$-3+i\sqrt{3} , 1+\sqrt{3}$

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

$\alpha = x+iy\\\beta = x-iy$

since they are conjugates

$\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}$

$\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}$

Imaginary part of the above =0

i.e. $\sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1$

So the value of alpha = $-3+i\sqrt{3} , 1+\sqrt{3}$

3 0

3 years ago

Write the equation for the function of the graph below​

Write the equation for the function of the graph below