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3 years ago

Due in 3 mins please help!!!

Mathematics

1 answer:

Korvikt [17]3 years ago

3 0

Answer:

a=100 sq. cm.

Step-by-step explanation:

a= l×w

a=10×10

a=100

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Proof: sin(x) + cos(x) = √2sin(x + π/4)

Oksi-84 [34.3K]

Hello :
sin(x) + cos(x) = √2(1/√2 sin(x) +1/√2 cos(x))
but : 1/√2 = cos(π/4) = sin(<span> π/4)
</span>sin(x) + cos(x) = √2( cos(π/4)sin(x) + sin( π/4) cos(x)) =√2<span>sin(x + π/4)
</span>because : cos(π/4)sin(x) + sin( π/4) cos(x)=sin(x + π/4) by identity :
sin(a+b) = sina cosb +cosa sinb

5 0

3 years ago

A large washer has an outer radius of 10mm and a hole with a diameter of 14mm. What is the area of the top surface of the washer

choli [55]

<h2>Answer:</h2>

The area of the top surface of the washer is: 160.14 square mm.

<h2>Step-by-step explanation:</h2>

The top of the surface is in the shape of a annulus with a outer radius of 10 mm and a inner radius of 7 mm ( since the diameter of the hole is: 14 mm and we know that the radius is half of the diameter)

Now, we know that the area of the annulus region is given by:

$Area=\pi (R^2-r^2)$

where R is the outer radius and r is the inner radius.

Here we have:

$R=10\ mm\\\\and\\\\r=7\ mm$

Hence, we have:

$Area\ of\ top\ surface=\pi (10^2-7^2)\\\\i.e.\\\\Area\ of\ top\ surface=\pi (100-49)\\\\i.e.\\\\Area\ of\ top\ surface=\pi\cdot 51\\\\i.e.\\\\Area\ of\ top\ surface=160.14\ mm^2$

4 0

3 years ago

Read 2 more answers

Solve for the unknown 3/15=9/x

Leni [432]

X = 3*15 = 45

Hope it helped.

3 0

2 years ago

PLEASE HELP ME BRAINLIEST WILL BE GIVEN!!

hjlf

Answer:

False

Step-by-step explanation:

The people arent random

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

2 years ago