Answer:
[2] x = -5y - 4
// Plug this in for variable x in equation [1]
[1] 2•(-5y-4) - 5y = 22
[1] - 15y = 30
// Solve equation [1] for the variable y
[1] 15y = - 30
[1] y = - 2
// By now we know this much :
x = -5y-4
y = -2
// Use the y value to solve for x
x = -5(-2)-4 = 6
Solution :
{x,y} = {6,-2}
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Terms and Topics
Linear Equations with Two Unknowns
Solving Linear Equations by Substitution
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Algebra - Linear Systems with Two Variables
Step-by-step explanation:
Answer:
$968
Step-by-step explanation:
x= the money John originally had in his savings account
The statement says that John spent $88 on a radio and 1/4 of what was left on presents for his friends. This means that the remaining money is 3/4 of the total amount minus the money spent on the radio:
3/4(x-88)
Then, it says that of the money remaining, John put 4/11 into a checking account and the last remaining $420 was left to charity. Since, he spent 4/11 of the remining 3/4(x-88), the remaining is 7/11 of that and that remaining is equal to $420 that was the money left at the end:
7/11(3/4(x-88))= 420
21/44(x-88)=420
21/44x-42=420
21/44x=462
x=968
According to this, the money that John originally had in his savings account was $968.
He spent $88 on a radio: 968-88= 880
1/4 of what was left on presents for his friends: 880/4= 220
880-220=660
Of the money remaining, John put 4/11 into a checking account: 660*(4/11)=240
the last remaining $420 was left to charity: 660-240= 420
Your answer would be 0,6.
- Because if you look at it correctly you see the right is at 0 and the top is at 6
Answer:
Subtract the two diameters and multiply by 0.5
Step-by-step explanation:
Calculating the Radius of the hole gives distance to the inner edges WHILE the Radius of the coin gives us the distance to the outer edges.
By subtracting the output we can then get the difference between both distances.
If hole diameter = 5mm and coin diameter = 22mm
Difference = [(22 - 5)mm] ÷ 2
17mm / 2 = 8.5mm
