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Usimov [2.4K]
3 years ago
13

Select the correct answer from each drop-down menu.

Mathematics
1 answer:
vazorg [7]3 years ago
6 0

Answer:

Step 1: Distribute -2 to 5x and 8

Step 2: Subtract from both sides of the equation 6x

Step 3: Add to both sides of the equation  16

Step 4: Divide both sides of the equation by -16

Step-by-step explanation:

Step 1: Apply the Distributive Property. Then you must  distribute -2 to 5x and 8

Then:

-2(5x+8)=14+6x\\\\-10x-16=14+6x

Step 2: You must apply the Subtraction property of Equality and subtract 6x from both sides of the equation. Then:

-10x-16-6x=14+6x-6x\\\\-16x-16=14

 Step 3: You must apply the Addition property of Equality and add 16 to both sides of the equation. Then:

-16x-16+16=14+16\\\\-16x=30

Step 4: You must apply the Division property of Equality and divide both sides by -16. Then:

\frac{-16x}{-16}=\frac{30}{-16}\\\\x=\frac{30}{-16}\\\\x=-\frac{15}{8}

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Nutka1998 [239]

Answer:

Vertex is (4,1)

Step-by-step explanation:

x=-b/2a so you need to first divide the b (second term) by the a (first term) so you should get 1 as your new a and -8 as your new b. Negative (-8) is just 8 then divide that by 2(1) which is just 2 which equals 4. That is your x. To find your y, you plug your x back into the original equation and then solve it. Hope this helps!

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3 years ago
What decimal is 0.99 less than 5
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3 years ago
Walt is mixing fruit punch for a party. He combines 1 gallon 2 quarts 3 pints of orange juice 1 gallon 3 quarts 7 pints of pinea
Vlad [161]
So he's mixing: 

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4 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
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