Need help extra points and mark brainlyist

Find the domain and range of the function y = f(x)= √x-2 <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7Bx%20-%202%7D%20" i

Over [174]

Answer:

3x-x+2=4

Step-by-step explanation:

The domain of f(x)=2^x would be the x values. This would include all values that you can input as x in order to make this problem work. The domain of a function is usually all real numbers. The range of f(x)=2^x would be the y values. This would include all values that would be the output for the y value. An example of this would be if you used 2 as x then the function would read f(x)=2^2. The y would equal 4 which would be included in the range of this function. To find the domain and range of the inverse you would follow the proper steps to get the inverse of the function which would be x=2^y. The domain would be the x values and the range would be the y values. If you put 4 as x which would be your input for the domain you would get 2^4 = 16 for the y which would be the range.

4 0

2 years ago

The number of chairs on a ski lift is inversely proportional to the distance between them a lift has 70 chairs when there’s spac

zhannawk [14.2K]

Answer:

Space between 80 evenly spaced chairs = 21m

Step-by-step explanation:

So, according to the given statement, we have the following expression:

No. of chairs ∝ $\frac{1}{Distance.Between.chairs}$

70 chairs are 24m apart, so it can be written as:

70 ∝ $\frac{1}{24}$

For, 1 chair , we divide both sides by 70

$\frac{70}{70}$ ∝ $\frac{1}{24(70)}$

1 ∝ $\frac{1}{24 (70)}$ -----------------------(Equation 1)

For 80 chairs:

Multiply both sides of Equation.1 by 80

80 ∝ $\frac{80}{24 (70)}$

80 ∝ $\frac{1}{21}$

So according to the above relation,

80 evenly spaced chairs will have a space of 21 m between them.

7 0

3 years ago

Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (4, −5, 2) and parallel

Nataliya [291]

Answer:

Step-by-step explanation:

From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e ( $x_o, y_o, z_o$ ) and are parallel to $\dfrac{x+5}{1} = \dfrac{y}{2}= \dfrac{z-3}{1}$

The parallel vector to the line i + zj+k = ai + bj + ck

Hence, the equation for the line is :

$x = x_o + at \\ \\ x = y_o + bt \\ \\ x = z_o + ct$

x = 4 + t

y = -5 + 2t

z = 2 + t

Thus, x, y, z = ( 4+t, -5+2t, 2+t )

The symmetric equation can now be as follows:

$\begin {vmatrix} x = 4+ t \\ \\ \dfrac{x-4}{1} = t \begin {vmatirx} \end {vmatrix}$ $\begin {vmatrix} y = - 5+2t \\ \\ \dfrac{y+5}{2} =t \end {vmatrix}$ $\begin {vmatrix} z =2+t \\ \\ \dfrac{z-2}{1} =t \end {vmatrix}$