Answer:
11+3=14
-41-1=-42
14-2(-42)
-2(-42)
14+84
98
I hope this is good enough:
So you want to find the price for 1 apple you find that by dividing so 23.60/5=4.72
and 32.76/7=<span>4.68
so, the 7 bag for 32.76 is the better deal</span>
There an old algorythmic method similar to long division:-
7 . 4 8 3
-----------------
) 56. 00 00 00
7 ) 49
) ---
144 ) 7 00
) 576
----
148 8 ) 12400
) 11904
-------
) 496 00
1496 3 ) 44889
THis gives the square root as 7.48 to the nearest hundredth
The formula for calculating length is:

We can also write

or

Why it does not matter?
Let's assume we have 2 numbers, a and b.
When we perform a subtraction:

, we get another number

When we perform another subtraction:

, we get a number

When we raise

or

to the power of 2, the result is the same,

.
Answer:
following are the solution to this question:
Step-by-step explanation:
Please find the complete question in the attached file.

The one-sided limits of F(x) at x = 3 must be equivalent for f(x) to be continuous at x = 3.
So,


In equation a multiply the by -2 and then add in the equation b:
So, the value of