All categories

3 years ago

Helppppppppp please

Mathematics

1 answer:

PolarNik [594]3 years ago

6 0

How there’s nothing to see we can’t see anything

You might be interested in

Write an equation for the line that passes through (2, -2) and (4, -1)

Dahasolnce [82]

Answer: y=1/2x-3

Step-by-step explanation:

Slope-intercept form:y=mx+b

m=y2-y1/x2-x1

-1-(-2)/4-2=1/2

y=1/2x+b

-1=1/2(4)+b

-1=2+b

b=-3

6 0

3 years ago

Each school bus going to the field trip holds 36 students and 4 adults . There are 6 filled uses on the field trip. How many peo

lukranit [14]

The answer is 240 please mark me as brainliest

7 0

3 years ago

What names apply to 50

sammy [17]

Do u mean like other names for it

5 0

3 years ago

The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in

Anon25 [30]

We have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is $[2,\infty)$ .

If we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

Let us know find its inverse.

$y=(x-5)^{2}+2$

Upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

Let us now solve this function for y.

$(y-5)^{2}=x-2\\
y-5=\pm \sqrt{x-2}\\
y=5\pm \sqrt{x-2}\\$

Hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

8 0

3 years ago

Write the quadratic function in standard form.<br><br> y=2(x - 3)² +9

xz_007 [3.2K]

Answer:

y = 2x^2 - 12x + 27

Step-by-step explanation:

<u>Step 1: Distribute the power</u>

y = 2(x - 3)² + 9

y = 2(x^2 - 6x + 9) + 9

y = 2x^2 - 12x + 18 + 9

y = 2x^2 - 12x + 27

Answer: y = 2x^2 - 12x + 27

5 0

3 years ago