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4 years ago

What is the answer? X/8 - 4 = 15/16

Mathematics

1 answer:

sukhopar [10]4 years ago

5 0

Answer:

Step-by-step explanation:

$\dfrac{x}{8}-4=\dfrac{15}{16}\qquad\text{multiply both sides by 8}\\\\8\!\!\!\!\diagup^1\cdot\dfrac{x}{8\!\!\!\!\diagup_1}-(8)(4)=8\!\!\!\!\diagup^1\cdot\dfrac{15}{16\!\!\!\!\!\diagup_2}\\\\x-32=\dfrac{15}{2}\qquad\text{add 32 to both sides}\\\\x-32+32=7.5+32\\\\x=39.5$

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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

Which of the following are solutions to the equation below (5x+1)^2=7?<br><br> Check all that apply.

stiv31 [10]

(5x+1)² = 7; expand: 25x² + 10x +1 = 7 or 25x² + 10x -6 = 0

Solve this quadratic for x:

x = [-b + √(b² - 4ac)]/2a and x = [-b - √(b² - 4ac)]/2a
Plug the values and you'll find :
x = (- √7 - 1)/5 (answer C)
x = (+√7 -1)/5 (answer E)

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3 years ago

Read 2 more answers

A soccer coach bought some new uniforms and cleats for his team.

natka813 [3]

Uhhhh i cant answer this without details

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3 years ago

Please be correct !!!!!!!!!!!!!!!!!xx

givi [52]

<span>$2,022.92 would be the answer

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