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Molodets [167]
3 years ago
11

Please send an explanation :)

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
7 0

Answer:

100 cm^2

Step-by-step explanation:

if FA and BF are the same and the shape is a square then  the right triangles

FBC and FEC have the same area. FA is 1/2 the length of the square.

if you rotate FEC to the bottom of the square the shaded are will be 1/2 of the square. 50 = 1/2 of the square... unshaded left square = 50 , and 50 = unshaded on the right

total unshaded should be 100 cm^2

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Colin pays £721.45 a year on his car insurance.
Burka [1]

Answer:

655.08

Step-by-step explanation:

<h3>721.45 ×0.092=66.3734</h3><h3>721.45-66.3734=655.0766</h3>
4 0
3 years ago
Which equation means the same as two more than the number of
mario62 [17]

Answer:

D. a + 2 = 5

Step-by-step explanation:

apples + 2 more = 5 total

8 0
3 years ago
Lisa recorded a 2-hour television show. When she watched it, she skipped the commercials. It took her
Alexxx [7]

Answer:

36 minutes

Step-by-step explanation:

2 hours in minutes is 120

120-84

36 minutes

8 0
3 years ago
A person can pay $33 for a membership to the history museum and then go to the museum
Lemur [1.5K]

Answer:

42$

Step-by-step explanation:

The person pays 33$ for a membership,and then visits the museum 9 times.So 33+(9*1)=42

4 0
3 years ago
Read 2 more answers
(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
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