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ohaa [14]
3 years ago
14

What's a multiplication equation that corresponds to this division equation 10÷5​

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
4 0

Answer:

2 × 5 = 10

Step-by-step explanation:

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Find the volume of a right circular cone that has a height of 16.8 in and a base with a
alexgriva [62]

Answer:

Volume of right circular cone is 388.43 in³

Step-by-step explanation:

Height of circular cone = 16.8 in

Radius of circular cone = 4.7 in

We need to find Volume of right circular cone

The formula used for calculating volume of a right circular cone is: Volume=\pi r^2\frac{h}{3}

Putting values and finding volume

Volume=\pi r^2\frac{h}{3}\\Volume=3.14\times (4.7)^2 \times \frac{16.8}{3}\\Volume=3.14\times 22.09 \times 5.6\\Volume= 388.43 \:in^3

So, Volume of right circular cone is 388.43 in³

5 0
2 years ago
The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to
Firdavs [7]

Rewrite the boundary lines <em>y</em> = -1 - <em>x</em> and <em>y</em> = <em>x</em> - 1 as functions of <em>y </em>:

<em>y</em> = -1 - <em>x</em>  ==>  <em>x</em> = -1 - <em>y</em>

<em>y</em> = <em>x</em> - 1  ==>  <em>x</em> = 1 + <em>y</em>

So if we let <em>x</em> range between these two lines, we need to let <em>y</em> vary between the point where these lines intersect, and the line <em>y</em> = 1.

This means the area is given by the integral,

\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy

The integral with respect to <em>x</em> is trivial:

\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy

For the remaining integral, integrate term-by-term to get

\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line <em>y</em> = 1 and (0, -1)), so its area is 1/2*4*2 = 4.

6 0
3 years ago
What is the value of z in the equation 2(4z − 5 − 3) = 166 − 46?
Vera_Pavlovna [14]
Z = 17.
2(4z -8) = 120
8z -16 = 120
z -2 = 15
z = 17
3 0
3 years ago
Read 2 more answers
A circle is growing so that the radius is increasing at the rate of 3 cm/min. How fast is the area of the circle changing at the
Naya [18.7K]

Answer:

The area is growing at a rate of \frac{dA}{dt} =226.2 \,\frac{cm^2}{min}

Step-by-step explanation:

<em>Notice that this problem requires the use of implicit differentiation in related rates (some some calculus concepts to be understood), and not all middle school students cover such.</em>

We identify that the info given on the increasing rate of the circle's radius is 3 \frac{cm}{min} and we identify such as the following differential rate:

\frac{dr}{dt} = 3\,\frac{cm}{min}

Our unknown is the rate at which the area (A) of the circle is growing under these circumstances,that is, we need to find  \frac{dA}{dt}.

So we look into a formula for the area (A) of a circle in terms of its radius (r), so as to have a way of connecting both quantities (A and r):

A=\pi\,r^2

We now apply the derivative operator with respect to time (\frac{d}{dt}) to this equation, and use chain rule as we find the quadratic form of the radius:

\frac{d}{dt} [A=\pi\,r^2]\\\frac{dA}{dt} =\pi\,*2*r*\frac{dr}{dt}

Now we replace the known values of the rate at which the radius is growing ( \frac{dr}{dt} = 3\,\frac{cm}{min}), and also the value of the radius (r = 12 cm) at which we need to find he specific rate of change for the area :

\frac{dA}{dt} =\pi\,*2*r*\frac{dr}{dt}\\\frac{dA}{dt} =\pi\,*2*(12\,cm)*(3\,\frac{cm}{min}) \\\frac{dA}{dt} =226.19467 \,\frac{cm^2}{min}\\

which we can round to one decimal place as:

\frac{dA}{dt} =226.2 \,\frac{cm^2}{min}

4 0
3 years ago
First, consider the number of moves as a
iren [92.7K]

Answer:

The sequence is 2, 4, 8, 16, 32

The next answer is 2

Step-by-step explanation:

4 0
2 years ago
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