D and E. If I remember right that should be the answer.
To check which ordered pair (point) is in the solution set of the system of given linear inequalities y>x, y<x+1; we just need to plug given points into both inequalities and check if that point satisfies both inequalities or not. If any point satisfies both inequalities then that point will be in solution.
I will show you calculation for (5,-2)
plug into y>x
-2>5
which is clearly false.
plug into y<x+1
-2<5+1
or -2<6
which is also false.
hence (5,-2) is not in the solution.
Same way if you test all the given points then you will find that none of the given points are satisfying both inequalities.
Hence answer will be "No Solution from given choices".
Answer:
x < 1
Step-by-step explanation:
an open circle means it contains no equal signs.....everything to the left is shaded....means it is less then
x < 1 (thats a less then sign only) ....no equal sign in there
Answer:
56 cm squared
Step-by-step explanation:
First things first: Cop one triangle off the rectangle, and attach it to the other one, so the shape looks like a L
Now I can actually solve this:
The left side is 8 cm (because y = 8 cm)
The top is 10 cm
The middle is 6 cm
The inside left is 3 cm
And the very bottom is 2 cm
First, we'll solve for the newly constructed rectangle: 3 x 2. That equals 6.
Next, solve for the longer rectangle: 10 x 5. That's 50.
Now, add the two areas, and we get 56. So the area of the whole thing is 56 cm squared.
(Please keep in mind that I could be wrong, so double check it for me, thanks!)
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π → C = π - (A + B)
→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))
→ sin C = sin (A + B) cos C = - cos(A + B)
Use the following Sum to Product Identity:
sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]
cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]
Use the following Double Angle Identity:
sin 2A = 2 sin A · cos A
<u>Proof LHS → RHS</u>
LHS: (sin 2A + sin 2B) + sin 2C
![\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]](https://tex.z-dn.net/?f=%5Ctext%7BFactor%3A%7D%5Cqquad%20%5Cqquad%20%5Cqquad%202%5Csin%20C%5Ccdot%20%5B%5Ccos%20%28A-B%29%2B%5Ccos%20%28A%2BB%29%5D)
LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C 
