Answer:
27/4 = 6.75
The final envelope will only contain 3 checks
Answer is 3
Explanation:
10 L -> 6 atm
So divide by half
5 L -> 3 atm
Answer:
Step-by-step explanation:
Given the simultaneous equation
2x - y = 11. Equation 1
6x – 3y = 15. Equation 2
Using elimination method
Multiply the coefficient of x in equation 1 (which is 2) to equation 2. Also, multiply the coefficient of x in equation 2 (which 6) to equation 1. By doing this we are going to eliminate x
2x - y = 11. × 6 Equation 1
6x – 3y = 15. ×2 Equation 2
12x-6y=66. Equation 3
12x-6y=30. Equation 4
Subtract equation 4 from 3
Then (12x-6y) - (12x-6y) =66 -30
0=36
Which is not possible
Then the system has no solution, it has no solution at all. Because 0 is not equal to 36. If has been 0 equals to 0, then we may many solutions for simultaneous equation.
<h3>30 dimes and 21 quarters are present</h3>
<em><u>Solution:</u></em>
Let "d" be the number of dimes
Let "q" be the number of quarters
1 dime = $ 0.10
1 quarter = $ 0.25
The worker counts 51 coins
Therefore,
d + q = 51
d = 51 - q ---------- eqn 1
<em><u>A parking meter contains $8.25 in dimes and quarters</u></em>
Therefore,
0.10d + 0.25q = 8.25 -------- eqn 2
<em><u>Substitute eqn 1 in eqn 2</u></em>
0.10(51 - q) + 0.25q = 8.25
5.1 - 0.10q + 0.25q = 8.25
0.15q = 3.15
Divide both sides by 0.15
<h3>q = 21</h3>
Substitute q = 21 in eqn 1
d = 51 - 21
<h3>d = 30</h3>
Thus 30 dimes and 21 quarters are present
Answer:
The missing side is 8 ( with radical form )
64 without doing radical
Step-by-step explanation:
Hope this helps
