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Volgvan
2 years ago
14

Can someone solve this question?​

Mathematics
1 answer:
iren2701 [21]2 years ago
5 0
It’s 26 bc it’s says it right there
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A group of students is painting the background of a set for a school play what is the area of the part that is painted in gray
GarryVolchara [31]
144 I’m not sure tho
6 0
3 years ago
Y = x + 2<br> 2x - y = -4
mrs_skeptik [129]
1.) X=-2

2.) 2x-y+4=0


Hope this helps! :))
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
An arch is in the shape of a parabola. It has a span of 100 feet and a maximum height of 25 ft.
faltersainse [42]

Answer:

y=-\frac{1}{100}(x-50)^2+25

the height of the arch 10 feet from the center is 24 feet

Step-by-step explanation:

An arch is in the shape of a parabola. It has a span of 100 feet, the vertex lies at the center 50 and the maximum height of 25 ft.

Vertex at (50,25)

vertex form of the equation is

y=a(x-h)^2+k, (h,k) is the center

y=a(x-50)^2+25

the parabola starts at (0,0) that is (x,y)

0=a(0-50)^2+25

subtract 25 from both sides

-25=a(-50)^2

-25=a(2500)

divide both sides by 2500

a=-\frac{1}{100}

y=-\frac{1}{100}(x-50)^2+25

the height of the arch 10 feet from the center.

center is at 50, 10 feet from the center so x=40 and x=60

y=-\frac{1}{100}(40-50)^2+25

y=24

the height of the arch 10 feet from the center is 24 feet

4 0
3 years ago
-1(x+2)=2(3x-6)<br> Please solve
Korvikt [17]
-1(x+2)=2(3x-6)
-1(x+2)=2(3x-6)distributhing -1 through parantheses
-1x+(-1)*2=2(3x-6)
-x-2=2(3x-6) distributhing 2 through parantheses
-x-2=2*3x-2*6
-x-2=6x-2*6
-x-2=6x-12    add x 
-2=6x-12+x
-2=7x-12        add 12
-2+12=7x
10=7x       divide by 7  
x= \frac{10}{7}
Answer : x= 1 3/7


6 0
3 years ago
Read 2 more answers
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