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Usimov [2.4K]
3 years ago
8

What is the surface area of this triangular right prism?

Mathematics
1 answer:
nignag [31]3 years ago
8 0
Area of triangular prism = 1/2 times base times height times length

1/2 * 5 * 6 * 11 (6.5 is extra info)
2.5 * 6 * 11
15 * 11
165 cubic feet or ft^3
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Please answer this now with correct answer
lana66690 [7]

Answer:

483.56 square milimeters

Step-by-step explanation:

In the above question, we obtain the following information:

Slant height = 15mm

Radius = 7mm

π = 3.14

Since we are given the slant height ,

the formula for surface area of a cone = πrl + πr²

= πr (l + r)

= 3.14 × 7(15 + 7)

= 3.14 × 7( 22)

= 21.98(22)

= 483.56 square milimeters

7 0
4 years ago
Robert bough 3/4 pound of grapes and divided them into 6 equal portions.
insens350 [35]

Answer: The answer should be 0.125

Step-by-step explanation:

4 0
3 years ago
Solve for x.<br><br><br>9x+4≤58 or 16x−2&gt;3<br><br><br> x 30<br><br> x 30
Brrunno [24]

Answer:

wow

Step-by-step explanation:

6 0
3 years ago
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What is the diameter of a cone with height 8 m and volume 150xm3?
Reika [66]
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6 0
4 years ago
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An isosceles trapezoid ABCD with height 2 units has all its vertices on the parabola y=a(x+1)(x−5). What is the value of a, if p
horrorfan [7]

Answer:

a = -0.3575

Step-by-step explanation:

The points A and D lie on the x-axis, this means that they are the x-intercepts of the parabola, and therefore we can find their location.

The points A and B are located where

y=a(x+1)(x-5)=0

This gives

x=-1

y=5

Now given the coordinates of A, we are in position to find the coordinates of the point B. Point B must have y coordinate of y=2 (because the base of the trapezoid is at y=0), and the x coordinate of B, looking at the figure, must be x coordinate of A plus horizontal distance between A and B, i.e

B_x=A_x+\frac{2}{tan(60)} =-1+\frac{2\sqrt{3} }{3}

Thus the coordinates of B are:

B=(-1+\frac{2\sqrt{3} }{3},2)

Now this point B lies on the parabola, and therefore it must satisfy the equation  y=a(x+1)(x-5).

Thus

2=a((-1+\frac{2\sqrt{3} }{3})+1)((-1+\frac{2\sqrt{3} }{3})-5)

Therefore

a=\frac{2}{((-1+\frac{2\sqrt{3} }{3})+1)((-1+\frac{2\sqrt{3} }{3})-5)}

\boxed{a=-0.3575}

8 0
3 years ago
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