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3 years ago

5

The quotient of x and 9 is greater than 27

1 answer:

Crazy boy [7]3 years ago

6 0

Yes it is (ima type cuz I need 20 characters )

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Is there some one who can help me please

lesya692 [45]

Answer:

I'd say $18

Step-by-step explanation:

I cant really tell, but your best guess is $18

6 0

3 years ago

X/5 + 3 = 9 <br> solve equation

Delicious77 [7]

Answer:

X= 15

Step-by-step explanation:

multiply 5 on x/5 and 9

divide 3 and 45

x=15

3 0

3 years ago

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

Solve the system of equations.<br> 2y+7x=−5<br> 5y−7x=12<br>

3241004551 [841]

We can solve this by substitution method.

Look at the second equation. If we rearrange to find 7x, we can substitute in the value into the first equation.

$5y-7x=12$

$5y-7x-12=0$

$5y-12=7x$

Therefore, $7x=5y-12$

Now replace the 7x in the first equation with 5y - 12:

$2y+7x=-5$ (substitute in 7x = 5y - 12)

$2y+(5y-12)=-5$

$7y-12=-5$

$7y=7$

$y=1$

Now that we know y, we can find x by substituting in y = 1 into any equation we want. I will use the equation: 7x = 5y - 12

$7x=5y-12$ (substitute in y = 1)

$7x=5(1) -12$

$5x=5-12$

$7x=-7$

$x=-1$

__________________________________________________________

<u>Answer:</u>

<u></u> $y=1\\x=-1$ <u></u>

4 0

3 years ago

What is the FCC of 12 and 30?<br> How to use the gcf to factor 12+30

Leno4ka [110]

Can i get time to answer this question.

3 0

3 years ago