Ok so this all about subtracting fractions this is what you want to do:
you can change 8 3/4 and 4 2/3 to have the same value and deominator
8 3/4 = 8 9/12(multiply both sides by 3)
4 2/3 = 4 8/12(multiply both sides by 4)
you do this because 12 is divisible by 3 and 4
then, subtract 4 8/12 from 8 9/12 because John is using some of the wood his dad gave him.
then subtract once again, taking 11/2 away from your difference.
again match the deominator of 11/2 to the deominator of your difference which should be 12. another divisible of 12 is 2 and 6.
11/2 = 66/12 = 5 6/12(multiply both the top and bottom by 6)
and start subtracting.
so John should have -1 5/12(?) wood left
Answer:
The answer is 200.96 cm^2.
2*3.14*r=50.24
=>r=50.24/6.28
=>r=8 cm
Thus area = 3.14*r*r
= 3.14*8*8
=200.96 cm^2
Step-by-step explanation:
The two lines in this system of equations are parallel
Step-by-step explanation:
Let us revise the relation between 2 lines
- If the system of linear equations has one solution, then the two line are intersected
- If the system of linear equations has no solution, then the two line are parallel
- If the system of linear equations has many solutions, then the two line are coincide (over each other)
∵ The system of equation is
3x - 6y = -12 ⇒ (1)
x - 2y = 10 ⇒ (2)
To solve the system using the substitution method, find x in terms of y in equation (2)
∵ x - 2y = 10
- Add 2y to both sides
∴ x = 2y + 10 ⇒ (3)
Substitute x in equation (1) by equation (3)
∵ 3(2y + 10) - 6y = -12
- Simplify the left hand side
∴ 6y + 30 - 6y = -12
- Add like terms in the left hand side
∴ 30 = -12
∴ The left hand side ≠ the right hand side
∴ There is no solution for the system of equations
∴ The system of equations represents two parallel lines
The two lines in this system of equations are parallel
Learn more:
You can learn more about the equations of parallel lines in brainly.com/question/8628615
#LearnwithBrainly
30/16=15/8...............
Answer:
x=24 y=33
Step-by-step explanation:
If a diameter or radius is perpendicular to a chord, then it bisects the chord and its arc.
