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Savatey [412]
3 years ago
13

HELP me please! Please I really need it!

Mathematics
1 answer:
Crank3 years ago
4 0

Answer:

The radius is 6.1 cm

Step-by-step explanation:

The volume of a sphere is

V = 4/3 pi r^3

There is a typo in the problem.

We know the volume

950 = 4/3 (3.14) r^3

Multiply each side by 3/4

3/4 *950 = 3.14 r^3

712.5 = 3.14 r^3

Divide each side by 3.14

712.5/3.14 = r^3

226.910828 = r^3

Take the cube root of each side

226.910828 ^ (1/3) = r^3 ^ 1/3

6.099371323 = r

The radius is 6.1 cm

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Answer:

There are different ways like the sides are proportional or two angles have the same measures in each triangle

Step-by-step explanation:

8 0
4 years ago
The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati
makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\

And the second one would be

y = 2-x \,\,\,\,\,,  1 \leq x \leq 2.

If you rotate the first region around the "y" axis you get that

{\displaystyle A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51 }

And if you rotate the second region around the "y" axis you get that

{\displaystyle A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188 }

And the sum would be  2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first  region around the x axis you get that

{\displaystyle A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708 }

And if you rotate the second region around the x axis you get that

{\displaystyle A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472 }

And the sum would be 1.5708+1.0472 = 2.618

7 0
4 years ago
Which phrase describes the variable expression z+ 8?
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Answer:

D

Step-by-step explanation:

In math increased means that you are adding.

Have a good day and a good rest of your week!

8 0
3 years ago
v={x,y,z} such that the points satisfies x-2y+3z=0. is v vector space? if not,find all for which v is not a victor space
atroni [7]

Yes, the set of vectors

V = {(x, y, z) : x - 2y + 3z = 0}

is indeed a vector space.

Let u = (x, y, z) and v = (r, s, t) be any two vectors in V. Then

x - 2y + 3z = 0

and

r - 2s + 3t = 0

Their vector sum is

u + v = (x + r, y + s, z + t)

We need to show that u + v also belongs to V - in other words, V is closed under summation. This is a matter of showing that the coordinates of u + v satisfy the condition on all vectors of V:

(x + r) - 2 (y + s) + 3 (s + t) = (x - 2y + 3z) + (r - 2s + 3t) = 0 + 0 = 0

Then V is indeed closed under summation.

Scaling any vector v by a constant c gives

cv = (cx, cy, cz)

We also need to show that cv belongs to V - that V is closed under scalar multiplication. We have

cx - 2cy + 3cz = c (x - 2y + 3z) = 0c = 0

so V is need closed under scalar multiplication.

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Four less than the product of twice a number and eight
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8+2*a-4 is the equation
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