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lyudmila [28]
3 years ago
15

I need help with this one to / surface area :)

Mathematics
2 answers:
nadezda [96]3 years ago
7 0

Step-by-step explanation:

the surface area (base area plus side wall) of a cone is, when we have the radius and the length of the slanted side wall :

SA = pi×r×(r + l)

in our case here

r = half of the diameter = 12/2 = 6 m

l = 13.4 m

so, we have

SA = pi×6×(6 + 13.4) = pi×6×19.4 = 365.6813849... m²

zzz [600]3 years ago
5 0

Answer:

389.9 to 1 decimal place

If you need another explanation comment on this one

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Solve 3y^{2} - (y + 2) (y - 2)
hram777 [196]

Answer:

3y^2 - (y + 2) (y - 2) = 0

<=> 3y^2 - (y^2 - 4) = 0

<=> 2y^2 + 4 =0

<=>y^2 + 2 = 0

=> Because y^2 is always equal or larger than 0, there is no real solution.

Hope this helps!

:)

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300,022,000.

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A car can travel 150 miles on 4 gallons of petrol. How far can it travel on 6 gallons?
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A boy visiting New York City views the Empire State building from a point on
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Answer:

49,820

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so what you were going to do first is you're going to do 940 * 53 there is which equals 2820 and then you're going to do turtle messes so then you're going to do 5 * 0 which equals zero then five times for which equals 20 + 5 * 9 which equals 45 + 2 which 45 + 2 equals 47 so you have 2820 + 47000 which equals 49,820

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A particle moves with velocity vector
asambeis [7]

Answer:

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-(2t+1)\hat{j}+(\frac{1}{3}t^3+1)\hat{k}

Step-by-step explanation:

We are given that velocity vector of a particle

\vec{v(t)}=t\hat{i}-2\hat{j}+t^2\hat{k}

When t=0 then the particle is at the point (0,-1,1).

We have to find the position of particle  at time t.

We know that

Velocity =\frac{Displacement }{time}=\frac{ds}{dt}

Therefore,\vec{v}=\frac{\vec{ds}}{dt}

\int{ds}=\int (t\hat{i}-2\hat{j}+t^2\hat{k})dt

Integrate on both sides then we get

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-2t\hat{j}+\frac{1}{3}t^3\hat{k}+C

\int x^n dx=\frac{x^{n+1}}{n+1}+C

Substitute the value of point at time t=0 then we get

C=-\hat{j}+\hat{k}

Substitute the value of C then we get

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-2t\hat{j}+\frac{1}{3}t^3\hat{k}-\hat{j}+\hat{k}

Therefore, the position of particle at time t

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-(2t+1)\hat{j}+(\frac{1}{3}t^3+1)\hat{k}

8 0
3 years ago
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