The answer is 50 because the formula for the area of a triangle is base x hight x 1/2. So if we plug both the 10s into the formula you get 10x10 which = 100 then divide that by 2 and you get 50. Hope it helps :)

5 0

3 years ago

Find the midpoint of the line segment with the endpoints A and B.

zavuch27 [327]

Answer:

Step-by-step explanation:

The midpoint M of two endpoints of a line segment can be found by using the formula

$M = ( \frac{x1 + x2}{2} , \: \frac{y1 + y2}{2} )\\$

From the question we have

$m = ( \frac{2 + 10}{2} \: , \: \frac{6 + 8}{2} ) \\ = (\frac{12}{2} \: , \: \frac{14}{2} ) \: \: \: \: \: \: \:$

We have the final answer as

Hope this helps you

3 0

3 years ago

maria took 7 hours to read a 259-page book and juanita took 5 hours to read a 175-page book. who read more pages per hour? expla

Temka [501]

Answer: Maria reads more pages per hour

Step-by-step explanation:

Maria - 37 pages per hour

259 = pages read

7 = hours

259/7

Juanita - 35 pages per hour

175 = pages read

5 = hours

175/5

3 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