Write an equation of the line that passes through the given point and has the given slope.

you have returned some merchandise to a store and received a store credit of $23. in the same store' you are purchasing picture

dmitriy555 [2]

Answer:

$40

Step-by-step explanation:

If you were to purchase 7 picture frames, you would replace x in the equation with 7, this is shown below.

f(x)=9x-23 ---> f(x)=9(7)-23

Now we can solve for f(x). First start with 9(7).

f(x)=9(7)-23 ---> f(x)=63-23

Now finish the last part by subtracting 23 from 63.

f(x)=63-23 ---> f(x)=40

This means that you would have to pay $40 dollars to purchase the 7 picture frames.

3 0

3 years ago

Is this a cylinder or prism

Flura [38]

prism

stupid work minimum

7 0

3 years ago

What is the coefficient of y3 in 5xy? 3x + 5y3 – 3?

poizon [28]

Answer:

Step-by-step explanation:

it is 5 because 5 is next to y ^3

8 0

4 years ago

If f(x) = 2x and g(x) = sqrt x,what is the domain of (fog)(x)?

Norma-Jean [14]

Answer:

All Real x larger or equal zero

or in inequality form: $x\geq 0$

or in interval notation: $[0,\infty)$

Step-by-step explanation:

If $f(x)=2x$ and $g(x)=\sqrt{x}$ , then using the definition for the composition of functions $(fo\,g)(x)=f(g(x))$ , we get:

$(fo\,g)(x)=f(g(x)) \\(fo\,g)(x)=f(\sqrt{x} )\\(fo\,g)(x)=2\,\sqrt{x}$

Base on this final expression, we understand that the Domain (numbers that can be used as input of the function and produce a real number as result) of this new function is limited to those x-values larger than or equal to zero. Negative numbers inside the square root will not produce a real number.

Therefore, the Domain is restricted to All Real numbers larger than or equal to o (zero).

This statement can also be given in inequality form as $x\geq 0$

or in interval notation: $[0,\infty)$

7 0

4 years ago

Find equations of the spheres with center(3, −4, 5) that touch the following planes.a. xy-plane b. yz- plane c. xz-plane

postnew [5]

Answer:

(a) (x - 3)² + (y + 4)² + (z - 5)² = 25

(b) (x - 3)² + (y + 4)² + (z - 5)² = 9

(c) (x - 3)² + (y + 4)² + (z - 5)² = 16

Step-by-step explanation:

The equation of a sphere is given by:

(x - x₀)² + (y - y₀)² + (z - z₀)² = r² ---------------(i)

Where;

(x₀, y₀, z₀) is the center of the sphere

r is the radius of the sphere

Given:

Sphere centered at (3, -4, 5)

=> (x₀, y₀, z₀) = (3, -4, 5)

(a) To get the equation of the sphere when it touches the xy-plane, we do the following:

i. Since the sphere touches the xy-plane, it means the z-component of its centre is 0.

Therefore, we have the sphere now centered at (3, -4, 0).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, -4, 0) as follows;

$d = \sqrt{(3-3)^2+ (-4 - (-4))^2 + (0-5)^2}$

$d = \sqrt{(3-3)^2+ (-4 + 4)^2 + (0-5)^2}$

$d = \sqrt{(0)^2+ (0)^2 + (-5)^2}$

$d = \sqrt{(25)}$

d = 5

This distance is the radius of the sphere at that point. i.e r = 5

Now substitute this value r = 5 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 5²

(x - 3)² + (y + 4)² + (z - 5)² = 25

Therefore, the equation of the sphere when it touches the xy plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 25

(b) To get the equation of the sphere when it touches the yz-plane, we do the following:

i. Since the sphere touches the yz-plane, it means the x-component of its centre is 0.

Therefore, we have the sphere now centered at (0, -4, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (0, -4, 5) as follows;

$d = \sqrt{(0-3)^2+ (-4 - (-4))^2 + (5-5)^2}$

$d = \sqrt{(-3)^2+ (-4 + 4)^2 + (5-5)^2}$

$d = \sqrt{(-3)^2 + (0)^2+ (0)^2}$

$d = \sqrt{(9)}$

d = 3

This distance is the radius of the sphere at that point. i.e r = 3

Now substitute this value r = 3 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 3²

(x - 3)² + (y + 4)² + (z - 5)² = 9

Therefore, the equation of the sphere when it touches the yz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 9

(b) To get the equation of the sphere when it touches the xz-plane, we do the following:

i. Since the sphere touches the xz-plane, it means the y-component of its centre is 0.

Therefore, we have the sphere now centered at (3, 0, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, 0, 5) as follows;

$d = \sqrt{(3-3)^2+ (0 - (-4))^2 + (5-5)^2}$

$d = \sqrt{(3-3)^2+ (0+4)^2 + (5-5)^2}$

$d = \sqrt{(0)^2 + (4)^2+ (0)^2}$

$d = \sqrt{(16)}$

d = 4

This distance is the radius of the sphere at that point. i.e r = 4

Now substitute this value r = 4 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 4²

(x - 3)² + (y + 4)² + (z - 5)² = 16

Therefore, the equation of the sphere when it touches the xz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 16

3 0

3 years ago