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Reptile [31]
3 years ago
6

Can somebody help me with this question

Mathematics
1 answer:
natta225 [31]3 years ago
4 0
First, we know that y is equal to -8x, so we can move the -8x to replace y in the first equation, then in the first equation we get -8x=6x-14, subtract 6x on both sides, -14x=-14, x=1
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Nikes cost $75.<br> They were reduced to $60. What is the percent of change?
Mamont248 [21]

Answer:

15%

75-60= 15

hope this helps

8 0
3 years ago
PLEASE HELP!!!<br> ANSWER ALL 4
wlad13 [49]

Answer:

1 and 8 are alternate exterior

8 and 5 are corresponding

3 and 2 are vertical

6 and 3 are alternate interior

4 0
2 years ago
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Scott is playing a game in which he rolls a number cube with faces numbered 1through 6 and spins the spinner shown below one tim
defon

Solution:

The total number of possiblities rolling a number cube with faces numbered 1 to 6 is;

And the total number of possibilities spining the spinner is;

The number of possibility where outcome on the cube is greater than 2 is;

4

And the number on the spinner less than 9 is;

2

Hence, the unique combination is;

4\times2=8

CORRECT OPTION: A

7 0
1 year ago
Which graph shows the solution to the following inequality? 4d-6&lt;18
Nadusha1986 [10]

Answer: C

<u>Step-by-step explanation:</u>

4d - 6 < 18

<u>      +6</u>   <u>+6 </u>

4d      < 24

    \frac{4d}{4} < \frac{24}{4}

       d < 6

Graph: ←----------------o 6

8 0
3 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
3 years ago
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