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2 years ago

Which inequality has infinitely many negative integer solutions

1 answer:

5 0

A. x>-5, that is greater than -5, so not this one
B. 5<x, that means x is bigger than 5, so not
C. x is greater than or equal to -10, so not this one
D. x is less than 10, so this one

D is answer

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For the second problem do I add, divide, subtract or multiplied

MA_775_DIABLO [31]

Divide hope it helps

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2 years ago

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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

2 years ago

You are choosing between two health clubs. Club A offers membership for a fee of $28 plus a monthly fee of $24. Club B offers me

Schach [20]

Answer:

The cost will be the same after 1 month

Step-by-step explanation:

Step 1: set up the equation

club A= 28x+24 club B= 35x+17

club A=club B: 28x+24=35x+17

Step 2: Get rid of the smallest X

28x+24=35x+17

-28 -28

24=7x+17

Step 3: Get rid of the constant next to the variable to isolate the variable

24=7x+17

-17 -17

7=7x

Step 4: Inverse operation to remove variable

7÷7x=7x÷7x

1=1

I hope this helped, im not too good at explaining but im positive the solution is correct.

3 0

3 years ago

Write the sentence as an equation z and 40 more is the same as 344

KengaRu [80]

Your answer z+40=344

3 0

1 year ago

If x is 5 what is the value of 4x?

Bond [772]

Answer:

The answer is 20

Step-by-step explanation:

This is because 4(5)= 20

Hope this helps :)

7 0

2 years ago

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