Find the smallest number that is divisible by 2, 3, 4, 5, 6 and add 1.
We need the least common multiple of 2, 3, 4, 5, 6.
2 = 2
3 = 3
4 = 2^2
5 = 5
6 = 2 * 3
LCM = product of common and not common prime factors with larger exponent.
LCM = 2^2 * 3 * 5 = 4 * 3 * 5 = 60
To always have a remainder of 1, you need of add 1 to 60.
The number is 61.
Check:
61/2 = 30 remainder 1
61/3 = 20 remainder 1
61/4 = 15 remainder 1
61/5 = 12 remainder 1
61/6 = 10 remainder 1
$12 x .20 = $2.40
12 - 2.40 = $9.60
$9.60 x .07 = $0.67
$9.60 + $0.67 = $10.27
Total price (including tax): $10.27
Step-by-step explanation (Question 1):
<u>Step 1: (Question 1): Subtract x from both sides.</u>
5x+3=x+13
5x+3−x=x+13−x
4x+3=13
<u>Step 2: (Question 1): Subtract 3 from both sides.</u>
4x+3−3=13−3
4x=10
<u>Step 3: (Question 1): Divide both sides by 4.</u>
4x/4 = 10/4
FIRST ANSWER: x = 5/2
Step-by-step explanation (Question 2):
<u>Step 1: (Question 2): Multiply by LCM</u>
15x^2 - 6 = x
<u>Step 2: (Question 2): Solve 15x^2</u>
15x^2 - 6 = x:
x = 2/3
x = -3/5
<u>Step 3: (Question 2): Solve</u>
ANSWER: x=0.666667 or x=−0.6
See Attachment 1 for question 1 steps (FULL)
See Attachment 2 for question 2 steps (FULL)
Answer:
1) x = 5/2
2) x=0.666667 or x=−0.6
Hope this helps.
Answer:
3x+15= 4x-10
3x-4x= -10-15
-x= -25(minus will be cut on both sides)
so x= 25
p q
| |
| | s
| ----------------------------------------------------- r
| | |
| | | -------------------------t
| | |
q || t is not true as you can see