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2 years ago

7.7 cm

Mathematics

1 answer:

svp [43]2 years ago

6 0

Step-by-step explanation:

I guess the 3.14 is to be used for pi (and not r, which is what we need to find out).

the volume of a cylinder is ground area × height.

the ground area is the area of a circle, which is pi×r².

so, we have

pi×r²×height = 115 cm³

3.14×r²×7.7 = 115 cm³

3.14×r² = 14.93506494...

r² = 4.756390107...

r = 2.18091497... cm

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Geometry help needed!

salantis [7]

Answer:

AC Midpoint (2,0)-

CB Midpoint (5,-4)

midlength is 5

Step-by-step explanation:

midpoint formula $(\frac{x_{1}+x_{2} }{2} ,\frac{y_{1} +y_{2} }{2})$

A(2,4) C(2,-4)

$\frac{2+2}{2} =\frac{4}{2}=2\\\frac{4-4}{2} =\frac{0}{2} =0$

(2,0)

C(2,-4) B(8,-4)

$\frac{8+2}{2} =\frac{10}{2} =5$

$\frac{-4-4}{2} =\frac{-8}{2} =-4$ midpoint is (5,-4)

Distance between the 2 midpoints is the midsegment

(2,0) and (5,-4)

$d=\sqrt{(x_{2}-x_{1})^{2} +(y_{2}-y_{1}) ^{2} }$

$d=\sqrt{(5-2)^{2} +(-4-0)^{2} } \\d=\sqrt{3^{2} +(-4^{2})\\ } \\d=\sqrt{x} 9+16\\d=\sqrt{25} \\d=5$

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3 years ago

Daniel rides his bicycle 21km west and then 18km north. How far is he from his starting point?

insens350 [35]

This is a right triangle. To find the length of the hypotenuse a^2 + b^2 = c^2.

21^2 + 18^2 = c^2

441 + 324 = c^2

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3 years ago

What is x^4+25 factored

OLga [1]

(x^2+5)(x^2-5)

I hope this helps :)

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3 years ago

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27 is 72% of what number?

NNADVOKAT [17]

Answer:

37.5

Step-by-step explanation:

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2 years ago

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Graph the hyperbola using the transverse axis, vertices, and co-vertices:

Reptile [31]

See attachment for the graph of the hyperbola 12x^2 - 3y^2 - 108 = 0

<h3>How to graph the hyperbola?</h3>

The equation of the hyperbola is given as:

12x^2 - 3y^2 - 108 = 0

Start by calculating the transverse axis

So, we have:

<u>Transverse axis</u>

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

Where

a = 3 and b = 6

The transverse axis is calculated as:

y = ±b/a(x - h) + k

So, we have:

y = ±6/3(x - 0) + 0

Evaluate the difference and sum

y = ±6/3x

Evaluate the quotient

y = ±2x

This means that the transverse axes are y = 2x and y =-2x

<u>The vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

<u>The co-vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

And

a = 3 and b = 6

The co-vertices are

(h - a, k) and (h + a, k)

So, we have:

(0 - 3, 0) and (0 + 3, 0)

Evaluate

(-3,0) and (3, 0)

See attachment for the graph of the hyperbola