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Yuki888 [10]
3 years ago
6

Maria claims that any fraction between 1/5 and 1/7 on a number line must have a denominator that is 6.Enter a fraction that show

s Maria's claim is incorrect A. 6/35 B.13/70 C. 9/35 D.16/70 please help me
Mathematics
1 answer:
Jet001 [13]3 years ago
8 0

Answer

A or B

Step-by-step explanation:

the lowest is 1/7 = 5/35 = 10/70

the highest is 1/5 = 7/35 = 14/70

so answer A and B are OK

C and D are out of range

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How many wholes are in 11/3?
lesantik [10]
There are 3 holes because if you do 4 thats to high cuz you would get 12 hope this is what you mean
5 0
2 years ago
At the beach 29% of people have red towels and 18% have white towels if the rest have blue ,what percentage of the people at the
Vikki [24]
The percent of people with blue towels would be the remaining percent when you add the red and wight towels together.
a.k.a
53% would have blue towels.


6 0
3 years ago
What is x? if u can’t see it the 2 other degrees are 32 and 120
Westkost [7]

Answer:28

Step-by-step explanation:

Since x+32 is 90 degrees, you subtract 32 from both sides to get the remaining which is 28 degrees.

8 0
3 years ago
(05.05)The width of a rectangle is shown below:
Ket [755]

Answer:

no idea sorry!!

Step-by-step explanation:

i think its the third but not sure

4 0
3 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
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