Answer:
5 length
Step-by-step explanation:
The diagram attached shows two equilateral triangles ABC & CDE. Since both squares share one side of the square BDFH of length 10, then their lengths will be 5 each. To obtain the largest square inscribed inside the original square BDFH, it makes sense to draw two other equilateral triangles AGH & EFG at the upper part of BDFH with length equal to 5.
So, the largest square that can be inscribe in the space outside the two equilateral triangles ABC & CDE and within BDFH is the square ACEG.
Answer:
(4,1)
Step-by-step explanation:
If you have any questions about the way I solved it, don't hesitate to ask me in the comments below :)
It either word for or number form
To get it into standard you need to simplify
6y - 12 = -3x add 12 to both sides
6y = -3x + 12 add 3x to both sides
3x + 6y = 12
And that is in standard which is A + B = C
Answer:
f(x) = -3|x| + 2x - 1
f(-5) = -3|-5| + 2*(-5) - 1
= -3*5 + 2*(-5) - 1
= -15 - 10 - 1
= -26
Hope this helps!
:)