You solve this by plugging one equation into the other. Usually you have to rewrite one equation to make this work. In this case I choose to rewrite y-4x=0 as y=4x.
After plugging it into the second, you get:
3x + 6*4x = 9 => 27x = 9 => x=1/3
Putting this solution back into y=4x gives us y=4/3
(5raise to power 5)÷(4)=3125/4)
Answer:
4:00 pm
Step-by-step explanation:
To find the time it takes Malik to finish his English essay, let's start by subtracting one hour.
5:30 minus 1 hour is 4:30.
Now, subtract 30 minutes.
4:30 minus 30 minutes is 4:00.
Malik started working on his English essay at 4:00 pm.
Hope that helps.
Answer:
20. AB = 42
21. BC = 28
22. AC = 70
23. BC = 20.4
24. FH = 48
25. DE = 10, EF = 10, DF = 20
Step-by-step explanation:
✍️Given:
AB = 2x + 7
BC = 28
AC = 4x,
20. Assuming B is between A and C, thus:
AB + BC = AC (Segment Addition Postulate)
2x + 7 + 28 = 4x (substitution)
Collect like terms
2x + 35 = 4x
35 = 4x - 2x
35 = 2x
Divide both side by 2
17.5 = x
AB = 2x + 7
Plug in the value of x
AB = 2(17.5) + 7 = 42
21. BC = 28 (given)
22. AC = 4x
Plug in the value of x
AC = 4(17.5) = 70
✍️Given:
AC = 35 and AB = 14.6.
Assuming B is between A and C, thus:
23. AB + BC = AC (Segment Addition Postulate)
14.6 + BC = 35 (Substitution)
Subtract 14.6 from each side
BC = 35 - 14.6
BC = 20.4
24. FH = 7x + 6
FG = 4x
GH = 24
FG + GH = FH (Segment Addition Postulate)
(substitution)
Collect like terms


Divide both sides by -3

FH = 7x + 6
Plug in the value of x
FH = 7(6) + 6 = 48
25. DE = 5x, EF = 3x + 4
Given that E bisects DF, therefore,
DE = EF
5x = 3x + 4 (substitution)
Subtract 3x from each side
5x - 3x = 4
2x = 4
Divide both sides by 2
x = 2
DE = 5x
Plug in the value of x
DE = 5(2) = 10
EF = 3x + 4
Plug in the value of x
EF = 3(2) + 4 = 10
DF = DE + EF
DE = 10 + 10 (substitution)
DE = 20
ill do as many as i can but i cant promise ill do them all
section 1:
(-13,-4) (-10,10) (-6,-13) (0,-14.5) (6,-13) (10,-11) (12,-8)