<span>Triangles exist:
</span>3 acute angles<span>
2 acute angles, 1 right angle
</span><span>2 acute angles, 1 obtuse angle
</span>
Triangles DON'T exist for:
<span>1 acute angle, 1 right angle, 1 obtuse angle
</span><span>1 acute angle, 2 obtuse angles</span>
The mean to your problem is 87.5
Here's what i did:
I broke down the fraction entirely.
3 / 5 = 6 / 10
divide 6 by two and divide 18 by two. This makes 3 / 10 = 9.
divide 3 by three and 9 by three, making 1 / 10 = 3.
if you need 7 / 10 of the number (30) then do 3*7.
The answer is 21.
Answer:
Step-by-step explanation:
dy/dx= -(2x+3y+1)/(3x-2y+1)
Let x= X+p. and y = Y+q. then dy/dx = dy/dX.
dy/dX= -(2X+2p+3Y+3q+1)/(3X+3p-2Y-2q+1) =
dy/dX = -{2X+3Y+(2p+3q+1)}/{3X-2Y+(3p-2q+1)}
Now 2p+3q+1=0……………..(1)
3p-2q +1=0…………………………(2)
p/(3+2)=q/(3–2)=1/(-4–9)
