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mote1985 [20]
2 years ago
6

A cone has height $h$ and a base with radius $r$ . You want to change the cone so its volume is doubled. What is the new height

if you change only the height
Mathematics
2 answers:
Viktor [21]2 years ago
3 0

\textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ V=\stackrel{doubled}{2V} \end{cases}\implies 2V=\cfrac{\pi r^2 h}{3}\implies V=\cfrac{\pi r^2 h}{3\cdot 2} \\\\\\ V=\cfrac{\pi r^2}{3}\cdot \cfrac{h}{2}\implies V=\cfrac{\pi r^2\left( \frac{h}{2} \right)}{3}\qquad \textit{half the previous "h"}

grandymaker [24]2 years ago
3 0

Answer: 2h

Step-by-step explanation:

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Select the equivalent expression.
Free_Kalibri [48]

Answer:

\frac{7^{15}}{3^{30}}

Step-by-step explanation:

The given expression is:

(\frac{3^{-6}}{7^{-3}})^{5}

Moving the expression to the other side in the fraction changes its sign to opposite. A numerator with negative exponent, when written in denominator will have the positive exponent. Using this rule, we can write:

(\frac{3^{-6}}{7^{-3}})^{5}\\\\ = (\frac{7^{3}}{3^{6}} )^{5}

The exponent 5 can be distributed to both numerator and denominator as shown:

(\frac{7^{3}}{3^{6}} )^{5}\\\\ = \frac{(7^{3})^{5}}{(3^{6})^{5}}

The power of a power can be written as a product. i.e.

\frac{(7^{3})^{5}}{(3^{6})^{5}}\\\\ =\frac{7^{15}}{3^{30}}

So, the expression similar to the given expression and with positive exponents is: \frac{7^{15}}{3^{30}}

6 0
3 years ago
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What's the digit in the tenths place ​
Darya [45]
Tenths is the digit in the tenths place. Like 0.1 being 1 tenth.
5 0
3 years ago
Which equation can be used to determine p, the cost of s swimsuits
ch4aika [34]

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6 0
3 years ago
On a linear X temperature scale, water freezes at −115.0°X and boils at 325.0°X. On a linear Y temperature scale, water freezes
belka [17]

Answer:

The current temperature on the X scale is 1150 °X.

Step-by-step explanation:

Let is determine first the ratio of change in X linear temperature scale to change in Y linear temperature scale:

r = \frac{\Delta T_{X}}{\Delta T_{Y}}

r = \frac{325\,^{\circ}X-(-115\,^{\circ}X)}{-25\,^{\circ}Y - (-65.00\,^{\circ}Y)}

r = 11\,\frac{^{\circ}X}{^{\circ}Y}

The difference between current temperature in Y linear scale with respect to freezing point is:

\Delta T_{Y} = 50\,^{\circ}Y - (-65\,^{\circ}Y)

\Delta T_{Y} = 115\,^{\circ}Y

The change in X linear scale is:

\Delta T_{X} = r\cdot \Delta T_{Y}

\Delta T_{X} = \left(11\,\frac{^{\circ}X}{^{\circ}Y} \right)\cdot (115\,^{\circ}Y)

\Delta T_{X} = 1265\,^{\circ}X

Lastly, the current temperature on the X scale is:

T_{X} = -115\,^{\circ}X + 1265\,^{\circ}X

T_{X} = 1150\,^{\circ}X

The current temperature on the X scale is 1150 °X.

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Paulina has read 58 pages of the 234-page book she is doing a book report on for Language Arts. She plans to finish her book by
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