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sattari [20]
3 years ago
14

Mary wants to fill in a cylinder vase. At the flower store they told her that the vase should be filled for the flowers to last

the longest. Her cylinder vase has a radius of 4in and a height of 10in. How much water should Mary pour into the vase?
Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
3 0
<span>The answer is 502,4 in^3. The volume of a cylinder (V) is π*r²*h, where r is the radius of the cylinder and h is the height of the cylinder. We know that π = 3.14, r = 4 in, h = 10 in. Substitute them in the formula for the volume of the cylinder: V = 3.14 * 4^2 * 10 = 3.14 * 16 * 10 = 502.4 in^3.</span>
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2 ^( 3 - 9) - 11 how do I solve this
Setler [38]
2^(3-9)-11 =

1 step: subtract the 3 and 9 to get 6

(3-9)=6

2 step: multiply 2 to its self 6 times and
you should get 64

2*2=4*2=8*2=16*2=32*2=64

3 step: subtract 11 by 64 to get 53

64-11=53

hope I help if so hit the brainliest thanks
5 0
3 years ago
Write the phrase as an expression. Then evaluate the expression when x = 5.
Nadusha1986 [10]

Answer:

6 + 8x

46

Step-by-step explanation:

1. Write the expression

6 + 8x

2. Plug in 5 for x

6 + 8(5) → 6 + 40 = 46

6 0
3 years ago
Write the quadratic equation in standard form. What is the value of b 2 - 4ac?
Fittoniya [83]
B^2 - 4ac > 0
2x^2 +7x -1 =0
b =7 a=2 c= -1
b^2 - 4ac = 7^2 - 4*2*(-1)
= 49 + 8
= 57
5 0
3 years ago
Help ASAP plz and put how to solve it
tatyana61 [14]

Answer: its 1/144

Step-by-step explanation:

3 0
3 years ago
Find the surface area of x^2+y^2+z^2=9 that lies above the cone z= sqrt(x^@+y^2)
Mashcka [7]
The cone equation gives

z=\sqrt{x^2+y^2}\implies z^2=x^2+y^2

which means that the intersection of the cone and sphere occurs at

x^2+y^2+(x^2+y^2)=9\implies x^2+y^2=\dfrac92

i.e. along the vertical cylinder of radius \dfrac3{\sqrt2} when z=\dfrac3{\sqrt2}.

We can parameterize the spherical cap in spherical coordinates by

\mathbf r(\theta,\varphi)=\langle3\cos\theta\sin\varphi,3\sin\theta\sin\varphi,3\cos\varphi\right\rangle

where 0\le\theta\le2\pi and 0\le\varphi\le\dfrac\pi4, which follows from the fact that the radius of the sphere is 3 and the height at which the sphere and cone intersect is \dfrac3{\sqrt2}. So the angle between the vertical line through the origin and any line through the origin normal to the sphere along the cone's surface is

\varphi=\cos^{-1}\left(\dfrac{\frac3{\sqrt2}}3\right)=\cos^{-1}\left(\dfrac1{\sqrt2}\right)=\dfrac\pi4

Now the surface area of the cap is given by the surface integral,

\displaystyle\iint_{\text{cap}}\mathrm dS=\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dv\,\mathrm du
=\displaystyle\int_{u=0}^{u=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}9\sin v\,\mathrm dv\,\mathrm du
=-18\pi\cos v\bigg|_{v=0}^{v=\pi/4}
=18\pi\left(1-\dfrac1{\sqrt2}\right)
=9(2-\sqrt2)\pi
3 0
3 years ago
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