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alexira [117]
3 years ago
10

The base of a triangular prism is a right triangle whose legs are 7 cm and 24 cm. The height of the prism is 30 cm. What is the

lateral area of the prism?
Mathematics
1 answer:
Grace [21]3 years ago
3 0
Ok, base area: triangle sides are 25cm each, or .25m
s=semiperimeter
area=s(sqrt(s-side)^3)
s=75/2
s-side=37.5-25=12.5
area=37.5*12.5*sqrt12.5
let E6 mean 10^6
.075E6m=1/3 (37.5*12.5*(sqrt12.5))h
solve for h
I get:137cm
I hope that helped

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notka56 [123]

Answer:

BE=15

Step-by-step explanation:

Since E is the center, BE=2x+1

And AC is 6x-12 with E being a midpoint.

Since AC is 2 segments and BE is only 1 then you multiply 2x+1 by 2 so that

you can find x. You multiply 2x+12 by 2 because E is the midpoint so that means ED is the same value as BE

6x-12=4x+2

BD=AC

6x-4x-12=2

2x=2+12

2x=14

x=7

BE= 2x+1

2(7)=14+1=15

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The Standard Form Is

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Factor completely. <br> <img src="https://tex.z-dn.net/?f=x%5E%7B8%7D-%5Cfrac%7B1%7D%7B81%7D" id="TexFormula1" title="x^{8}-\fra
Eduardwww [97]

We have 3⁴ = 81, so we can factorize this as a difference of squares twice:

x^8 - \dfrac1{81} = \left(x^2\right)^4 - \left(\dfrac13\right)^4 \\\\ x^8 - \dfrac1{81} = \left(\left(x^2\right)^2 - \left(\dfrac13\right)^2\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)

Depending on the precise definition of "completely" in this context, you can go a bit further and factorize x^2-\frac13 as yet another difference of squares:

x^2 - \dfrac13 = x^2 - \left(\dfrac1{\sqrt3}\right)^2 = \left(x-\dfrac1{\sqrt3}\right)\left(x+\dfrac1{\sqrt3}\right)

And if you're working over the field of complex numbers, you can go even further. For instance,

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But I think you'd be fine stopping at the first result,

x^8 - \dfrac1{81} = \boxed{\left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)}

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3 years ago
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