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lina2011 [118]
2 years ago
5

Alica had 10/12 of fabric she used 8/12 for a pillow how much fabric dod she have left? Explain how you found your answer .

Mathematics
1 answer:
Gala2k [10]2 years ago
5 0
If she has 10/12 and uses 8/12 then you just subtract.
Since the denominators are the same you just subtract 10-8 and get 2
the denominator is the same for your answers and so it is 
2/12
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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)
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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

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