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Tpy6a [65]
3 years ago
12

Factorize 216x^3 + 64y^3

Mathematics
1 answer:
marin [14]3 years ago
5 0

Answer:

8( 3x+2y) ( 9x^2 -6xy +4y^2)

Step-by-step explanation:

216x^3 + 64y^3

Factor out the greatest common factor of 8

8( 27x^3 +8y^3)

Rewriting

8 ( (3x)^3 + (2y)^2))

Recognizing this as the difference of cubes)

a^3 + b^3 = (a+b) (a^2-ab+b^2)

8 ( (3x)^3 + (2y)^2)) = 8( 3x+2y) ( 9x^2 -(3x)(2y) +4y^2)

                                 =8( 3x+2y) ( 9x^2 -6xy +4y^2)

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Given △ABC. m∠A>m∠B>m∠C. Perimeter=30. Which side of △ABC may have length 7?
lisabon 2012 [21]

<u>Solution-</u>

As given in △ABC,

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As from the properties of trigonometry we know that, the greater the angle is, the greater is the value of its sine. i.e

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According to the sine law,

\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}

In order to make the ratio same, even though m∠A>m∠B>m∠C, a must be greater than b and b must be greater than c.

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Also given that its perimeter is 30. Now we have to find out whose side length is 7. So we have 3 cases.

Case-1. Length of a is 7

As a must be the greatest, so b and c must be less than 7. Which leads to a condition where its perimeter won't be 30. As no 3 numbers less than 7 can add up to 30.

Case-2. Length of b is 7

As b is greater than c, so c must 6 or less than 6. But in this case the formation of triangle is impossible. Because the triangle inequality theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side. If b is 7 and c is 6, then a must be 17. So no 2 numbers below 7 can add up to 17.

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As this is the last case, this must be true.

Therefore, by taking the aid of process of elimination, we can deduce that side c may have length 7.


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3 years ago
A bag of flour weighed 2.25 kilograms 0.725 kilograms are used in a recipe how many kilograms of flour are left
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Hello!

You subtract the amount of flour you started with by the amount you used

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Hope this helps!
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a = interest rate of first CD

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\bf X=X\qquad thus\qquad \implies \cfrac{24000}{a}=\cfrac{36000}{b}\implies \cfrac{24000}{a}=\cfrac{36000}{\boxed{3+a}} \\\\\\ (3+a)24000=36000a\implies \cfrac{3+a}{a}=\cfrac{36000}{24000}\implies \cfrac{3-a}{a}=\cfrac{3}{2} \\\\\\ 6-2a=3a\implies 6=5a\implies \cfrac{6}{5}=a\implies 1\frac{1}{5}=a\implies \blacktriangleright 1.2 = x\blacktriangleleft

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