How to solve your problem
(4−3)(−22−7−5)
(4x−3)(−2x2−7x−5)(4x-3)(-2x^{2}-7x-5)(4x−3)(−2x2−7x−5)
Simplify
1
Distribute
(4−3)(−22−7−5)
(4x−3)(−2x2−7x−5){\color{#c92786}{(4x-3)(-2x^{2}-7x-5)}}(4x−3)(−2x2−7x−5)
4(−22−7−5)−3(−22−7−5)
2
distribute
4(−22−7−5)−3(−22−7−5)
4x(−2x2−7x−5)−3(−2x2−7x−5){\color{#c92786}{4x(-2x^{2}-7x-5)}}-3(-2x^{2}-7x-5)4x(−2x2−7x−5)−3(−2x2−7x−5)
−83−282−20−3(−22−7−5)
3
−83−282−20−3(−22−7−5)
−8x3−28x2−20x−3(−2x2−7x−5)-8x^{3}-28x^{2}-20x{\color{#c92786}{-3(-2x^{2}-7x-5)}}−8x3−28x2−20x−3(−2x2−7x−5)
−83−282−20+62+21+15
4
Solution
−83−222++15
I know this looks like a lot but its just how you solve your problem.
therefor, your answer is
Solution
−83−222++15
<em>Hope this helps!</em>
<em>Have a great day!</em>
<em>-Hailey</em>
We know from Euclidean Geometry and the properties of a centroid that GC=2GM. Now GM=sin60*GA=
. Hence CM=GC+GM=3*
. Now, since GM is normal to AB, we have by the pythagoeran theorem that:
Hence, we calculate from this that AC^2= 28, hence AC=2*
=BC. Thus, the perimeter of the triangle is 2+4*
.
Answer: See explanation
Step-by-step explanation:
We are given the numbers 5,4,2, and 2 and told to make an equation from it and get a final answer as 24. This will be:
= 5 × 4 + 2 + 2
= 20 + 2 + 2
= 24.
Since we have gotten a final answer of 24, the equation is solved.
