Answer:
The inequality sign remains same while multiply or divide both sides by positive numbers.
The inequality sign changes while multiply or divide both sides by negative numbers.
Step-by-step explanation:
The given inequality is - 8 < 2.
Now, if we multiply 2 in both sides then - 16 < 4
Again, if we divide by 2 into both sides then - 4 < 1
Therefore, the inequality sign remains the same while multiply or divide both sides by positive numbers.
Now, if we multiply -2 in both sides then 16 > -4
And, if we divide -2 into both sides then 4 > -1
Therefore, the inequality sign changes while multiply or divide both sides by negative numbers. (Answer)
Answer: 0.03269754768
Step-by-step explanation:
divide 12/367
search it up
Answer:
A two-digit number can be written as:
a*10 + b*1
Where a and b are single-digit numbers, and a ≠ 0.
We know that:
"The sum of a two-digit number and the number obtained by interchanging the digits is 132."
then:
a*10 + b*1 + (b*10 + a*1) = 132
And we also know that the digits differ by 2.
then:
a = b + 2
or
a = b - 2
So let's solve this:
We start with the equation:
a*10 + b*1 + (b*10 + a*1) = 132
(a*10 + a) + (b*10 + b) = 132
a*11 + b*11 = 132
(a + b)*11 = 132
(a + b) = 132/11 = 12
Then:
a + b = 12
And remember that:
a = b + 2
or
a = b - 2
Then if we select the first one, we get:
a + b = 12
(b + 2) + b = 12
2*b + 2 = 12
2*b = 12 -2 = 10
b = 10/2 = 5
b = 5
then a = b + 2= 5 + 2 = 7
The number is 75.
And if we selected:
a = b - 2, we would get the number 57.
Both are valid solutions because we are changing the order of the digits, so is the same:
75 + 57
than
57 + 75.
Answer:
$19,747.96
Step-by-step explanation:
You are going to want to use the continuous compound interest formula, which is shown below:
<em>A = total</em>
<em>P = principal amount</em>
<em>r = interest rate (decimal)</em>
<em>t = time (years)</em>
<em />
First, lets change 5.5% into a decimal:
5.5% -> 
-> 0.055
Next, plug in the values into the equation:
After 5 years, you will have $19,747.96
3/7 is the answer for this question
