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LenaWriter [7]
3 years ago
6

V = s^2 + 1/2sh ; solve for h please provide answer & clear explanation

Mathematics
1 answer:
Elan Coil [88]3 years ago
7 0

Answer:

h = 2 (v − s²) / s

Step-by-step explanation:

v = s² + ½sh

Subtract s² from both sides.

v − s² = ½sh

Multiply both sides by 2.

2 (v − s²) = sh

Divide both sides by s.

2 (v − s²) / s = h

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Mary bought 3 shirts. The price of each shirt was $20.53. How much did the 3 shirt cost? This is 23 point question please I need
Olenka [21]

Answer:

$61.59

Step-by-step explanation:

1 shirt = $20.53

3 shirts x $20.53 = $61.59

to check:

$61.59 ÷ 3 shirts = $20.53 each

4 0
3 years ago
The 2007 record for the University of North Carolina softball team was 46 wins to 21
Oksana_A [137]

Answer:

the ratio is still 46:21 because it cn't be reduced any further

5 0
3 years ago
I NEED HELP ON THIS PROBLEM!!! PLEASE
Allushta [10]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

5x+8y=-9\implies 8y=-5x-9\implies y=\cfrac{-5x-9}{8} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{5}{8}} x-9\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}

well then, so since this equation has that slope therefore

\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{-5}{8}} ~\hfill \stackrel{reciprocal}{\cfrac{8}{-5}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{8}{-5}\implies \cfrac{8}{5}}}

so we're really looking for the equation of a line whose slope is 8/5 and runs through (10,10)

(\stackrel{x_1}{10}~,~\stackrel{y_1}{10}) ~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{8}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{\cfrac{8}{5}}(x-\stackrel{x_1}{10})

5 0
2 years ago
Suppose astronomers built a 50-meter telescope. how much greater would its light-collecting area be than that of the 10-meter ke
drek231 [11]

Answer: Its light-collecting area would be 25 times greater than that of the 10-meter keck telescope.

Step-by-step explanation:

1. To solve this problem you must apply the formula for calculate the area of a circle, which is shown below:

A=r^{2}\pi

Where<em> r</em> is the radius of the circle.

2. The diameter of a circle is:

D=\frac{r}{2}

Where <em>r</em> is the radius of the circle.

3. Therefore, keeping this on mind, you have that the light-collecting area of a  50-meter keck telescope is:

A_1=(\frac{50m}{2})^{2}\pi=1963,49m^{2}

4. And the light-collecting area of a 10-meter keck telescope is:

A_2=(\frac{10m}{2})^{2}\pi=78.53m^{2}

5. Divide A_1 by A_2, then:

\frac{1963.49m^{2}}{78.53m^{2}}=25

6. Therefore, its light-collecting area would be 25 times greater than that of the 10-meter keck telescope.

4 0
4 years ago
Kelon decides to put a garden with a fence aro-<br> und it, how much fencing<br> is needed? ?
11111nata11111 [884]

Answer:

What are the dimensions of the garden?

Step-by-step explanation:

8 0
3 years ago
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