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pantera1 [17]
3 years ago
10

Will a stand that can hold up to 40 pounds support a 21 kilogram television

Mathematics
2 answers:
Eduardwww [97]3 years ago
7 0
Covert 21kg to pound
0.45kg = 1 pound
21 kg would be equal to;
=\frac{21*1}{0.45}
=46.2971 pounds
The stand cannot support the stand
shtirl [24]3 years ago
5 0
If you would like to know if a stand will support a 21-kg television, you can find this using the following steps:

1 kilogram = <span>2.20462262 pounds
1 kilogram = 2.2 pounds

21 kilogram = 21 * 1 kilogram = 21 * 2.2 pounds = 46.2 pounds
46.2 pounds > 40 pounds

The correct result would be: No, the stand won't support a television.</span>
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Step-by-step explanation:

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Pleasee help anyone please
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Answer:

x must be 14/4 = 7/2

Step-by-step explanation:

The sum of the three interior angles of a triangle must be 180 degrees.

Thus, in this particular case,

82 + 64 + 4x + 20 = 180, or

   146     + 4x = 160, or

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\overleftrightarrow{AC} AC A, C, with, \overleftrightarrow, on top is tangent to circle OOO at point CCC. What is the length of
IgorC [24]

Answer:

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Step-by-step explanation:

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F ind the volume under the paraboloid z=9(x2+y2) above the triangle on xy-plane enclosed by the lines x=0, y=2, y=x
olga nikolaevna [1]

Answer:

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Step-by-step explanation:

If we simply draw out the region on the x-y plane enclosed between these lines we realize that,if we evaluate the integral the limits all in all cannot be constants since one side of the triangular region is slanted whose equation is given by y=x. So the one of the limit of one of the integrals in the double integral we need to evaluate must be a variable. We choose x part of the integral to have a variable limit, we could well have chosen y's limits as non constant, but it wouldn't make any difference. So the double integral we need to evaluate is given by,

V=\int\limits^2_0 {} \, \int\limits^{x=y}_0 {z} \, dx dy\\V=\int\limits^2_0 {} \, \int\limits^{x=y}_0 {9(x^{2}+y^{2})} \, dx dy

Please note that the order of integration is very important here.We cannot evaluate an integral with variable limit last, we have to evaluate it first.after performing the elementary x integral we get,

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After performing the elementary y integral we obtain the desired volume as below,

V= 48 units^{3}

4 0
3 years ago
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