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2 years ago

If a triangle being has an area of 10 square feet what is the length and width of the cover

Mathematics

1 answer:

bekas [8.4K]2 years ago

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Question needs elaboration.

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Let T be the triangular region in the xy plane with vertices (0,0),(B,0),(B,H), where B,H are positive constants. Use a double i

Llana [10]

Answer:

Step-by-step explanation:

Find the solution below

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3 years ago

The sides of two cubes are in the ratio 2:1. What is the ratio of their volumes ?

olasank [31]

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Answer:

we know that, Volume of cube=a³

Let volume of 1st cube be 2x with side a and other be x with side A (according to given ratio)

Step-by-step explanation:

ATQ: 2x=a³ and, x=A³

a=³√2x and, A=x

we know that, surface area of cube is 6a²

Thus, surface area of 1st cube = 6(³√2x)²

= 6³√4x²

Surface area of 2nd cube=6(x)²=6x²

Ratio of S.A=(6³√4x²)÷(6x²)

Ratio of S.A=³√4:1

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2 years ago

What is 720° converted to radians?

deff fn [24]

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2 years ago

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Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$