Help me find the arc

Starting at home, Luis traveled uphill to the hardware store for 3030 minutes at just 88 mph. He then traveled back home along t

dybincka [34]

Answer:

the distance from home to hardware store = (3030/60) * 88= 4444miles

time to travel back at a speed of 24mph = 4444miles /24mph = 11110 minutes

7 0

3 years ago

PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ HELP MEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

fenix001 [56]

Answer:

a

Step-by-step explanation:

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

Please I don’t understand help me?!

prohojiy [21]

Answer:

BD = 35

Step-by-step explanation:

Calculate CD in right triangle ABC, then BD in right triangle BCD

Using Pythagoras' identity in both triangles.

The square on the hypotenuse is equal to the sum of the squares on the other two sides.

In Δ ADC

CD² + AD² = AC² , substitute values

CD² + 9² = 15²

CD² + 81 = 225 ( subtract 81 from both sides )

CD = 144 ( take the square root of both sides )

CD = $\sqrt{144}$ = 12

----------------------------------------------------------------

In Δ BCD

BD² + CD² = BC² , substitute values

BD² + 12² = 37²

BD² + 144 = 1369 ( subtract 144 from both sides )

BD² = 1225 ( take the square root of both sides )

BD = $\sqrt{1225}$ = 35

5 0

3 years ago

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60 POINTS! (Reported if incomplete/absent)

GuDViN [60]

Answer:

- The neighbor that has the most amount of fence to put up is: Neighbor A.

- The neighbor that has the least amount of fence to put up is: Neighbor C.

Step-by-step explanation:

To calculate the amount of fence that Neighbor A has to put up, you need to add the following distances:

Convert yards to feet (Remember that :

To calculate the amount of fence that Neighbor B has to put up, you need to use the formula for calculate the circumference of a circle:

Where r is the radius.

In this case, the radius is:

Substituting, you get:

To calculate the amount of fence that Neighbor C has to put up, you need to add the following distances:

Therefore, the neighbor that has the most amount of fence to put up is: Neighbor A.

The neighbor that has the least amount of fence to put up is: Neighbor C.

hope this helps btw i just copy and pasted this so its not my work

4 0

3 years ago

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