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AURORKA [14]
3 years ago
5

PLSS I NEED HELP I NEED HELP SOMEONE SAVE ME

Mathematics
1 answer:
S_A_V [24]3 years ago
8 0

Answer:

sorry but are you dyin why do u need help why do you need someone to save you just say i need answers to this equation pls

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A conical paper cup has a radius of 2 inches. Approximate, to the nearest degree, the angle β so that the cone will have a volum
dezoksy [38]
Volume = area of base *height 

20 = pir^2 *h

20 = 3,14*4*h

h = 20/12,56 = 1,59 rounded 1,6

- using the triangle formed by radius,the height of cup and the slant height of cup so we can writing that

tan ß = h/r so tan ß = 1,6/2 = 0,8

tan ß = 0,8 so ß = arctan 0,8 = 38,6 degrees so 39 degrees 

hope helped 
5 0
3 years ago
What is the square root of 50 to the nearest tenth
krok68 [10]
7.07 so I think you'd leave it at 7 for your answer
7 0
3 years ago
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The volume of a cube 512 cubic inches. What is the length of the cube?
aleksandr82 [10.1K]

Answer:

I believe it is 8in

Step-by-step explanation:

someone correct me if I'm wrong

8 0
3 years ago
Graph the quadratic functions y=-2xandy=-2x2 + 4 on a separate piece of paper. Using those graphs, compare and contrast the shap
dedylja [7]

Y = -2x^2

h = Xv = -B/2A = 0/-4 = 0.

k = -2*0^2 = 0.

V(0,0).

Use the following points for graphing:

Y = -2x^2.

(x,y)

(-2,-8)

(-1,-2)

V(0,0)

(1,-2)

(2,-8)

Y = -2x^2 + 4.

h = Xv = -B/2A = 0/-4 = 0.

k = Yv = -2*0^2 + 4 = 4.

V(0,4).

(-2,-4)

(-1,2)

V(0,4)

(1,2)

(2,-4)

8 0
2 years ago
The base of a solid is the region enclosed by the graphs of y=e^x, y=0, x=0, and x = 1. If each cross-section perpendicular to t
yanalaym [24]

Answer:

A

Step-by-step explanation:

The base of a solid in the region enclosed by the graphs of <em>y</em> = eˣ, <em>y</em> = 0, <em>x </em>= 0, and <em>x</em> = 1. Each cross-section perpendicular to the <em>x</em>-axis is an equilateral triangle. We want to find the volume of the solid.

Please refer to the graph below. We are concerned with the red region.

In order to find the volume, we essentially sum up the area of the figure at each <em>x</em> value. So, we integrate from <em>x </em> = 0 to <em>x </em>= 1.

The area for an equilateral triangle is given by:

\displaystyle A=\frac{\sqrt{3}}{4}s^2

Where <em>s</em> is the side length of the triangle.

Since the triangle lies perpendicular on the region, the side length of the triangle at <em>x</em> is simply <em>y</em>, which is eˣ.

Therefore, our volume is:

\displaystyle V=\int_0^1\frac{\sqrt3}{4}(y)^2\, dx

Substitute:

\displaystyle V=\int_0^1\frac{\sqrt3}{4}(e^x)^2\, dx

Evaluate the integral. Simplify:

\displaystyle V=\frac{\sqrt3}{4}\int_0^1e^{2x}\, dx

Integrate using u-substitution:

\displaystyle V=\frac{\sqrt3}{8}\left(e^{2x}\Big|_0^1\right)

Evaluate:

\displaystyle V=\frac{\sqrt3}{8}\left(e^{2(1)}-e^{2(0)} \right)

Therefore, the volume of the solid is:

\displaystyle V=\frac{\sqrt3}{8}\left(e^2-1\right)

Our answer is A.

3 0
3 years ago
Read 2 more answers
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