Solve with the substitution method -6x+4=33 5x+18y=29
1 answer:
You have to solve for 1 variable
I'm pretty sure they want 4y and not just y becaseu taht is too easy
-6x+4y=33
5x+18y=29
solve for 1 variable
may be messy
-6x+4y=33
add 6x both sides
4y=33+6x
divide both sides by 4
y=(33+6x)/4
sub that for y in other equation
5x+18((33+6x)/4)=29
5x+9(33+6x)/2=29
5x+(297+54x)/2=29
(10x/2)+(297+54x)/2=29
times both sides by 2
10x+297+54x=58
297+64x=58
minus 297 both sides
64x=-239
divide both sides by 64
x=-239/64
sub back
y=(33+6x)/4
y=(33+6(-239/64))/4
y=(33-717/32)/4
y=(1056/32-717/32)/4
y=(339/32)/4
y=339/128
the solution is
you can graph it and check if you want
I'm pretty sure they want 4y and not just y becaseu taht is too easy
-6x+4y=33
5x+18y=29
solve for 1 variable
may be messy
-6x+4y=33
add 6x both sides
4y=33+6x
divide both sides by 4
y=(33+6x)/4
sub that for y in other equation
5x+18((33+6x)/4)=29
5x+9(33+6x)/2=29
5x+(297+54x)/2=29
(10x/2)+(297+54x)/2=29
times both sides by 2
10x+297+54x=58
297+64x=58
minus 297 both sides
64x=-239
divide both sides by 64
x=-239/64
sub back
y=(33+6x)/4
y=(33+6(-239/64))/4
y=(33-717/32)/4
y=(1056/32-717/32)/4
y=(339/32)/4
y=339/128
the solution is
you can graph it and check if you want
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Answer:
Answer:
1) Factor form :
2) 8 second after launch.
Step-by-step explanation:
The height of the ball (in meters above the ground) t seconds after launch is modeled by
To find the time when ball hit the ground, we need to find the factor form of the given function.
When ball hi the ground, then height of the ball from the ground is 0.
Using zero product property, we get
Ball hit the ground at t=0 and t=8. It means ball hit the ground in starting and 8 second after launch.
Check the picture below.
bear in mind that a reference angle, is just the angle the "terminal point" makes with the x-axis, in this case, the green one.
bear in mind that a reference angle, is just the angle the "terminal point" makes with the x-axis, in this case, the green one.