Answer:
There are a few solutions because there are some fractions and decimals between 8 and 10
Step-by-step explanation:
Let the unknown number be 'x'
If the number is greater than 8 and the same number is less than 10, this can be expressed as;
x>8 and x < 10
Note that if x>8, then 8<x
The resulting inequalities are now;
8<x and x<10
Combining both inequalities we have: 8<x<10
Since the inequality didn't tell us that the variable 'x' is equal to 8 and 10, this means that our solution falls between 8 and 10 and the value of integer that falls within this range is 9. Other values that falls within this range are decimals and fractions.
Therefore it can be concluded that there are a few solutions because there are some fractions and decimals between 8 and 10
Answer:
16
−81
Step-by-step explanation:
Hope this is correct and it helps :)
Step-by-step explanation:
8⁴.... I hope it's correct
HAVE A GOOD DAY!
Answer:
Step-by-step explanation:
a = 2, b = 1, c = -3
We need to factor this by finding the product of a and c, then from there find which factors of a * c will either add or subtract to give us b.
a * c = 6 and the factors of 6 and 1 and 6, 2 and 3. Well, 6 - 1 doesn't equal 1 and neither does 6 + 1. So our factors are 3 and 2. In order to combine those to get a 1 (our b), we will subtract 2 from 3 since 3 - 2 = 1. That means that 3 is positive and 2 is negative. Filling in the formula with 3 and 2 in place of 1 looks like this (always remember to put the absolute value of the largest number first):

Group the first 2 terms together and the second 2 term together in order to factor:
and factor out what's common in each set of parenthesis.

Notice that when we factor out a -1 from the second set of parenthesis, we can distribute it back in to get the equation we started with. We know that factoring by grouping "works" if what is inside both sets of parenthesis is exactly the same. Ours are identical: (2x + 3). That is common now, and can be factored out:

That matches your first choice
The answer would be 2. You have to do multiplication first.