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andreev551 [17]

3 years ago

10

Show twelve in four different ways

1 answer:

nlexa [21]3 years ago

4 0

Ok i hope this is the answer that you are looking for
6*2=12one way
3×4=12second way
12×1=12third way
Sorry if this is not what you are looking for and I couldn't find the fourth way!!
Or incase there is another ways like 6+6=12,9+3=12,12+0=12,24÷2=12
Sorry if I did this wrong

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Use the identity (x2+y2)2=(x2−y2)2+(2xy)2 to determine the sum of the squares of two numbers if the difference of the squares of

fomenos

Answer:

The sum of the squares of two numbers whose difference of the squares of the numbers is 5 and the product of the numbers is 6 is <u>169</u>

Step-by-step explanation:

Given : the difference of the squares of the numbers is 5 and the product of the numbers is 6.

We have to find the sum of the squares of two numbers whose difference and product is given using given identity,

$(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2$

Since, given the difference of the squares of the numbers is 5 that is $(x^2-y^2)^2=5$

And the product of the numbers is 6 that is $xy=6$

Using identity, we have,

$(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2$

Substitute, we have,

$(x^2+y^2)^2=(5)^2+(2(6))^2$

Simplify, we have,

$(x^2+y^2)^2=25+144$

$(x^2+y^2)^2=169$

Thus, the sum of the squares of two numbers whose difference of the squares of the numbers is 5 and the product of the numbers is 6 is 169

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3 years ago

Read 2 more answers