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3 years ago

Put this equation in slope intercept form 5x-2y=10

Mathematics

1 answer:

Katen [24]3 years ago

4 0

This is the answer and the steps for doing it

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Please assist me in 6-30

Ymorist [56]

Answer:

Yes to both.

Step-by-step explanation:

It is true that

$\dfrac{x}{y}=1 \iff x=y$

In fact, if you know that x/y=1, just multiply both sides by y to get x=y. On the other hand, if you know that x=y (and they are not zero), you can divide both sides by y to get x/y=1.

Since the two expressions are equivalent, you can always use Phil's or Don's expression, at will.

8 0

3 years ago

F(x)=3x+5, g(x)=6x^2 find (fg)(x)

alexgriva [62]

Answer:

18x^3+30x^2

Step-by-step explanation:

im smart.

6 0

3 years ago

Which numbers are solutions

Tom [10]

Answer:

6 and 3

Step-by-step explanation:

x^2 - 9x + 18 = 0

(x-6)•(x-3)=0

if x-6 = 0 then x = 6

and if x-3= 0 then x = 3

3 0

3 years ago

Read 2 more answers

If y varies directly with x and k = 3, what will y equal when x = 2?

Nikolay [14]

Y=3x
x-2
The correct answer is y=6

5 0

3 years ago

Read 2 more answers

Find the standard equation of a sphere that has diameter with the end points given below. (3,-2,4) (7,12,4)

DiKsa [7]

Answer:

The standard equation of the sphere is $(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

Step-by-step explanation:

From the question, the end point are (3,-2,4) and (7,12,4)

Since we know the end points of the diameter, we can determine the center (midpoint of the two end points) of the sphere.

The midpoint can be calculated thus

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Let the first endpoint be represented as $(x_{1}, y_{1}, z_{1})$ and the second endpoint be $(x_{2}, y_{2}, z_{2})$ .

Hence,

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Midpoint = $(\frac{3 + 7 }{2}, \frac{-2+12 }{2}, \frac{4 + 4 }{2})$

Midpoint = $(\frac{10 }{2}, \frac{10}{2}, \frac{8 }{2})\\$

Midpoint = $(5, 5, 4)$

This is the center of the sphere.

Now, we will determine the distance (diameter) of the sphere

The distance is given by

$d = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2} + (z_{2}- z_{1})^{2} }$

$d = \sqrt{(7 - 3)^{2} +(12 - -2)^{2} + (4- 4)^{2}$

$d = \sqrt{(4)^{2} +(14)^{2} + (0)^{2}$

$d = \sqrt{16 +196 + 0$

$d =\sqrt{212}$

$d = 2\sqrt{53}$

This is the diameter

To find the radius, r

From $Radius = \frac{Diameter}{2}$

$Radius = \frac{2\sqrt{53} }{2}$

∴ $Radius = \sqrt{53}$

r = $\sqrt{53}$

Now, we can write the standard equation of the sphere since we know the center and the radius

Center of the sphere is $(5, 5, 4)$

Radius of the sphere is $\sqrt{53}$

The equation of a sphere of radius r and center $(h,k,l)$ is given by

$(x-h)^{2} + (y-k)^{2} + (z-l)^{2} = r^{2}$

Hence, the equation of the sphere of radius $\sqrt{53}$ and center $(5, 5, 4)$ is

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = \sqrt{(53} )^{2}$

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

This is the standard equation of the sphere

6 0

3 years ago