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3 years ago

The height of a cone is twice the radius of its base.

Mathematics

1 answer:

lara [203]3 years ago

6 0

Answer:

$A. \frac{2}{3} \pi x^3$

Step-by-step explanation:

From the way the answers are presented, it can be seen that x refers to the radius of the base of the cone

radius: $x$

and we are told that the height is twice the radius, so:

height: $2x$

and now we use the formula to calculate the volume of a cone:

$V=\frac{\pi r^2h}{3}$

where $V$ is volume, $r$ is radius, and $h$ is the height. and $\pi$ is a constant

in this case

$r=x$

$h=2x$

so we substitute thisvalues in the formula for the volume:

$V=\frac{\pi x^2(2x)}{3}$

Rearranging the terms

$V=\frac{2\pi x^3}{3} \\V=\frac{2}{3} \pi x^3$

which is option A.

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Find the area of the figure sides meet at right angles 8 cm, 3cm, 4cm, 5cm, 9cm

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Answer:

umm is it a quadrilateral

Step-by-step explanation:

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3 years ago

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If Mary Alice buys the car for $35,000 and pays 20% down, how much will she finance? If she finances this balance for 6 years at

RUDIKE [14]

Answer:monthly is 598.90 total is 43120

Step-by-step explanation:

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3 years ago

Express 25km/h into m/s?

Helen [10]

Answer:

3600(seconds) can also be expressed as 1(kilometer/hour) = 5/18 (meters/second), which is its simplified form. To convert km/h to m/s, directly multiply the given value of speed by the fraction 5/18.

Step-by-step explanation:

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2 years ago

(c). It is well known that the rate of flow can be found by measuring the volume of blood that flows past a point in a given tim

aleksklad [387]

(i) Given that

$V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr$

when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then

$V(0.30) = \displaystyle \int_0^{0.30} 2\pi \left(0.30-3.33r^2\right)r \, dr \approx \boxed{0.0425}$

and this is a volume so it must be reported with units of cm³.

In Mathematica, you can first define the velocity function with

v[r_] := 0.30 - 3.33r^2

and additionally define the volume function with

V[R_] := Integrate[2 Pi v[r] r, {r, 0, R}]

Then get the desired volume by running V[0.30].

(ii) In full, the volume function is

$\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

Compute the integral:

$V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

$V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr$

$V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R$

$V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right)$

$V(R) = \displaystyle \boxed{\frac{\pi KR^4}2}$

In M, redefine the velocity function as

v[r_] := k*(R^2 - r^2)

(you can't use capital K because it's reserved for a built-in function)

Then run

Integrate[2 Pi v[r] r, {r, 0, R}]

This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with

Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]

which ensures that R is positive, and moreover a real number.

5 0

3 years ago

Help please with this question

choli [55]

Answer:

They lose about 2.79% in purchasing power.

Step-by-step explanation:

Whenever you're dealing with purchasing power and inflation, you need to carefully define what the reference is for any changes you might be talking about. Here, we take <em>purchasing power at the beginning of the year</em> as the reference. Since we don't know when the 6% year occurred relative to the year in which the saving balance was $200,000, we choose to deal primarily with percentages, rather than dollar amounts.

Each day, the account value is multiplied by (1 + 0.03/365), so at the end of the year the value is multiplied by about

... (1 +0.03/365)^365 ≈ 1.03045326

Something that had a cost of 1 at the beginning of the year will have a cost of 1.06 at the end of the year. A savings account value of 1 at the beginning of the year would purchase one whole item. At the end of the year, the value of the savings account will purchase ...

... 1.03045326 / 1.06 ≈ 0.9721 . . . items

That is, the loss of purchasing power is about ...

... 1 - 0.9721 = 2.79%

_____

If the account value is $200,000 at the beginning of the year in question, then the purchasing power <em>normalized to what it was at the beginning of the year</em> is now $194,425.14, about $5,574.85 less.

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3 years ago