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3 years ago

a triangular sail has a base length of 2.5 meters. the area of the sail is 3.75 square meters. how tall is the sail?

Mathematics

2 answers:

PSYCHO15rus [73]3 years ago

7 0

Knowing that are of a triangle = 1/2 base * height this would make the area 0.5*5*10 or 25sq m

Alik [6]3 years ago

4 0

Area=1/2 times height times base
base=2.5
area=3.75
subsitute

3.75=1/2 times height times 2.5
multiply both sides by 2 to get rid of fraction
7.5=height itmes 2.5
divide both sides by 2.5
3=height

sail is 3 meters tall

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For every 10 flowers,there are 36 butterflies resting on the flower with the same number on each flower .How many butterflies ar

Gemiola [76]

Answer:

I don't know if this is right or not, but here's what I think.

There are 36 butterflies on each flower.

Step-by-step explanation:

Question:

For each 10 flowers, there are 36 butterflies resting on the flower with the same number on each flower. How many butterflies are on one flower?

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3 years ago

Show that 5(3y-2)=15y-10 for at least 5 values of y

Vsevolod [243]

Solving the equation for Y = 1. 5 * (3 * 1 - 2) = 15 * 1 - 10 5 * (3 - 2) = 15 -10 5 * 1 = 5 5 = 5 Solving the equation for Y = 2. 5 * (3 * 2 - 2) = 15 * 2 - 10 5 * (6 - 2) = 30 -10 5 * 4 = 20 20 = 20 Solving the equation for Y = 4. 5 * (3 * 4 - 2) = 15 * 4 - 10 5 * (12 - 2) = 60 -10 5 * 10 = 50 50 = 50 Solving the equation for Y = 8. 5 * (3 * 8 - 2) = 15 * 8 - 10 5 * (24 - 2) = 120 -10 5 * 22 = 110 110 = 110 Solving the equation for Y = 9. 5 * (3 * 9 - 2) = 15 * 9 - 10 5 * (27 - 2) = 135 -10 5 * 25 = 125 125 = 125 This proves that the equation holds good for at least 5 values of 'y', which are 1, 2, 4, 8 and 9. However, it can be proved that the equation holds good for any value of y. Expression 5(3y-2) can be simplified to 15y -10 which is the same expression on the right had side of the equation provided. So, equation 5(3y-2)=15y-10 is actually 15y-10=15y-10 and since this is true for all values of y, it has been proved that it is true for at least 5 values of y.

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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.

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maks197457 [2]

Answer:

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Step-by-step explanation:

basically you just plug in all of the x values shown in the table into the equation, to find the matching y values.

i hope this helps :)

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cluponka [151]

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Hope this helps you!

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3 years ago