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What are the range values of the function f (x ) = 3x + 4 for the domain values {-5, -1, 0, 2, 7}?

Airida [17]

Answer:

Range values are:

{-11, 1, 4, 10, 25}

Step-by-step explanation:

Domain values are the possible x values that can be entered into the function, or the inputs. Range values are the y values, or the outputs. By inserting every value in the set into the function, we get the outputs (y values)

-15 + 4 = -11

-3+4 = 1

0 + 4 = 4

6 + 4 = 10

21 + 4 = 25

Remember to put them in order.

7 0

3 years ago

Someone help me with this

Alenkasestr [34]

Answer:

70°

Step-by-step explanation:

ABC is the center angle that sees the arc and is twice as angle ADC so the measure of ABC is e × 35 = 70

3 0

4 years ago

Read 2 more answers

Write the equation of a circle that has a center of (-8, 0)<br> and a diameter of 8.

Sveta_85 [38]

Answer:

(x+8)^2+y^2=16

Step-by-step explanation:

The equation for a circle in center-radius form is

(x-h)^2+(y-k)^2=r^2

where (h,k) is the center and r is the radius.

We are given the diameter is 8 so the radius is 8/2=4.

We are also given (h,k) is (-8,0).

The equation for the circle is

(x--8)^2+(y-0)^2=4^2

(x+8)^2+y^2=16

8 0

3 years ago

Assuming that the equation defines x and y implicitly as differentiable functions xequalsf(t), yequalsg(t), find the slope of

Doss [256]

Answer:

$\dfrac{dx}{dt} = -8,\dfrac{dy}{dt} = 1/8\\$

Hence, the slope , $\dfrac{dy}{dx} = \dfrac{-1}{64}$

Step-by-step explanation:

We need to find the slope, i.e. $\dfrac{dy}{dx}$ .

and all the functions are in terms of $t$ .

So this looks like a job for the 'chain rule', we can write:

$\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx} -Eq(A)$

Given the functions

$x = f(t)\\y = g(t)\\$

and

$x^3 +4t^2 = 37 -Eq(B)\\2y^3 - 2t^2 = 110 - Eq(C)$

we can differentiate them both w.r.t to $t$

first we'll derivate Eq(B) to find dx/dt

$x^3 +4t^2 = 37\\3x^2\frac{dx}{dt} + 8t = 0\\\dfrac{dx}{dt} = \dfrac{-8t}{3x^2}\\$

we can also rearrange Eq(B) to find x in terms of t , $x = (37 - 4t^2)^{1/3}$ . This is done so that $\frac{dx}{dt}$ is only in terms of t.

$\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\$

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

$\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\\dfrac{dx}{dt} = \dfrac{-8(3)}{3(37 - 4(3)^2)^{2/3}}\\\dfrac{dx}{dt} = -8$

now let's differentiate Eq(C) to find dy/dt

$2y^3 - 2t^2 = 110\\6y^2\frac{dy}{dt} -4t = 0\\\dfrac{dy}{dt} = \dfrac{4t}{6y^2}$

rearrange Eq(C), to find y in terms of t, that is $y = \left(\dfrac{110 + 2t^2}{2}\right)^{1/3}$ . This is done so that we can replace y in $\frac{dy}{dt}$ to make only in terms of t

$\dfrac{dy}{dt} = \dfrac{4t}{6y^2}\\\dfrac{dy}{dt}=\dfrac{4t}{6\left(\dfrac{110 + 2t^2}{2}\right)^{2/3}}\\$

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

$\dfrac{dy}{dt} = \dfrac{4(3)}{6\left(\dfrac{110 + 2(3)^2}{2}\right)^{2/3}}\\\dfrac{dy}{dt} = \dfrac{1}{8}$

Finally we can plug all of our values in Eq(A)

but remember when plugging in the values that $\frac{dy}{dt}$ is being multiplied with $\frac{dt}{dx}$ and NOT $\frac{dx}{dt}$ , so we have to use the reciprocal!