Step-by-step explanation:
I'm going to assume you meant to write
in the equation as +1 - 1 wouldn't make much sense since they would just cancel out.
a. 
b. 
c. 
Answer:
No Problem
Step-by-step explanation:
I couldn’t find you on Inst. M’I missing something?
The line needs to be 8.06 m long
Explanation
Given that the ladder is 7 feet tall and the bottom of the slide should be 4 m away from the ladder.
The ladder,slide and the ground connecting them makes the sides of a right angled triangle with ladder as the height and ground as the base.
base=4m
height=7m
hypotenuse can be determined using pythagoras law

Answer:
The solution is x=1,y=2,z=3
Step-by-step explanation:
The given system of equations is ;\
2x−3y+4z=8...(1)
3x+4y−5z=−4...(2)
4x−5y+6z=12...(3)
Make x the subject in equation (1)

Put equation (4) into equation (2) and (3)

Multiply through by;

Expand;

Simplify;

Equation (4) in (3)





Put equation (6) into equation (5)




z=3
Put z=3 into equation (6)
y=2(3)-4=2
Put y=2 and z=3 into equation 4

The solution is x=1,y=2,z=3