The scale on a map is 1cm equals 25mi. How many miles would be 3cm be equivalent to?

There is 1 inch of water in a pool. The water level is increasing at 0.75 inches/minute. Which linear equation represents the to

diamong [38]

The answer is y = 0.75x + 1

4 0

2 years ago

What is the slope of a line perpendicular?<br><br> y=-3/4x +3

almond37 [142]

4/3
Remember if two lines are perpendicular, the product of their slopes are -1

8 0

3 years ago

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triangle MNO is an equilateral triangle with sides measuring 16 units What is the height of the triangle?

fiasKO [112]

see the attached figure to better understand the problem

we know that

The equilateral triangle has three equal sides

so

in the equilateral triangle ABC

$AB=BC=AC=16 units$

the height of the triangle is the segment BD

in the right triangle BCD

Applying the Pythagorean Theorem

$BC^{2} =BD^{2}+DC^{2}$

solve for BD

$BD^{2}=BC^{2}-DC^{2}$

substitute the values

$BD^{2}=16^{2}-8^{2}$

$BD^{2}=192$

$BD=\sqrt{192}\ units$

therefore

<u>the answer is</u>

the height of the triangle is $\sqrt{192}\ units$

7 0

3 years ago

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(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

Several ordered pairs from a continuous exponential function are shown in the table. what are the domain and range of function?

son4ous [18]

the answer is b......

5 0

3 years ago

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