D. 192in^2
This would be a simple area problem with a triangle. REMEMBER THIS EQUATION: a=bh*1/2 (b for base and h for height, these are multiplied together then that answer is halved out.)
So we just need to plug in our values into the equation, so the equation would look like a= (16)*(24)*1/2. 16 times 24 would then give you 384, you could either divide by 2 or multiply by 0.5 to get the next answer, as long as your HALVING the answer.
So we have our bh value so now we can multiply by 1/2 which will give us 384*1/2 which leaves us with 192.
I have also attached a photo of doing a longer(ish) way than this, that also proves that this equation works. Either one will provide you an answer.
The answer is 4x2-5x;Area= 114 in<span>2 i think</span>
Answer:
Step-by-step explanation:
(x^2+y^2)^2=(x^2)^2+2x^2y^2+(y^2)^2
Adding and substracting 2x^2y^2
We get
(x^2+y^2)^2=(x^2)^2+2x^2y^2+(y^2)^2 +2x^2y^2-2x^2y^2
And we know a^2-2ab+b^2=(a-b)^2
So we identify (x^2)^2 as a^2 ,(y^2)^2 as b^2 and -2x^2y^2 as - 2ab. So we can rewrite (x^2+y^2)^2=(x^2 - y^2)^2 + 2x^2y^2 + 2x^2y^2= (x^2 - y^2)^2+4x^2y^2= (x^2 - y^2)^2+2^2x^2y^2
Moreever we know (a·b·c)^2=a^2·b^2·c^2 than means 2^2x^2y^2=(2x·y)^2
And (x^2+y^2)^2=(x^2 - y^2)^2 + (2x·y)^2