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Leokris [45]
3 years ago
6

Geometry help! this is my 2nd to last question

Mathematics
2 answers:
Kryger [21]3 years ago
7 0
A square since all sides are equal and the angles aren't acute so it can't be a rhombus
Scrat [10]3 years ago
3 0
Hi.
Your answer would be d. square. Rhombus have actute angles, squares have right angles. It's clearly not a rectangle (as the sides are all the same) so it has to be square.

I hope I helped and if you need more you could always ask me :)

-Dawn
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Lady_Fox [76]
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4 0
3 years ago
What is the perimeter of a rectangle with sides 3 ft and (c+4) ft?
Alika [10]

Answer:

(2c+14) ft

Step-by-step explanation:

The answer is (2c+14) ft because as 2 sides measure 3 feet, we have 6 feet for two sides. The other two sides measure (c+4) feet, and there is (2c+8) feet for two sides. If you add the results together, you get (2c+14) feet. Hope it helps!

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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.

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gogolik [260]
The answer to this question is 32
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Elis [28]

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