Answer:
Step-by-step explanation:
you gotta simplify it,

After simplifying, your answer would be 4-12x
Hope this helps
Hello,
s=48t-16t²
a)
s=0==>16t(3-t)=0==>t=0 or t=3
b)
s>32==>48t-16t²>32===>16(t²-3t+2)<0
Δ=9-8=1
==>16(t-1)(t-2)<0
==>1<t<2 (negative between the roots)
Answer:
hello attached below is the detailed solution
answer : In |x+1| + [2/(x+1)] + [1/(1+x)^2] - [y^2/2(1+x)^2] = 5/2
Step-by-step explanation:
Given
(x^2 + y^2 - 3) dx = ( y + xy ) dy, y(0) = 1
solving the given initial-value problem
Here you are trying to find the hypotenuse. So you would take
5^2 (five squared) plus 12^2 (twelve squared) and add them using the equation a^2+b^2= c^2. Now you square 5 and 12 to get 25 and 144. You add them together and get 169. Take the square root of 169 and c. The square root on c will cancel out the squared (to the power of two) part leaving you with just c. And the square root of 169 is 13. So c (the missing length) is 13. C=13.