Solution:
(6y^2-9y+4)-(-7y^2+5y+1)
Distribute to get rid of ( )
6y^2-9y+4+7y^2-5y-1
Combine like terms
13y^2-14y+3
They didn't make it easy. Grid lines are apparently 3 apart, but the offered coordinates are all multiples of 4. It appears the only point that is in the doubly-shaded area is ...
... B (-4, -10)
30 bags: each pound can make 5 bags (1/5) so 6 x 5 = 30
See the picture attached to better understand the problem
we know that
in the right triangle ABC
cos A=AC/AB
cos A=1/3
so
1/3=AC/AB----->AB=3*AC-----> square----> AB²=9*AC²----> equation 1
applying the Pythagoras Theorem
BC²+AC²=AB²-----> 2²+AC²=AB²---> 4+AC²=AB²----> equation 2
substitute equation 1 in equation 2
4+AC²=9*AC²----> 8*AC²=4----> AC²=1/2----> AC=√2/2
so
AB²=9*AC²----> AB²=9*(√2/2)²----> AB=(3√2)/2
the answer isthe hypotenuse is (3√2)/2
Answer:
0
Step-by-step explanation:
The answer is 0 due to using PEMDAS