One simple way to approach this is to subtract the number that DOESN’T have the variable with the solution by the sum.
Lets use problem 1 as the example:
1.) N+8=13
In this case, we subtract 13 by 8 and get 5, therefore N= 5.
2.) B+17=62
62-17= 45
B= 45
3.) X+5/6= 7/16
7/16-5/6= -19/48
X= -19/48
4.) y+1/7=5/8
5/8-1/7= 27/56
Y= 27/56
5.) 14+y=69
69-14=55
Y=55
6.) c+3.6=4.9
4.9-3.6=1.3
C= 1.3
Now for the word problems:
Lets use 7 as the example-
7.) a number increased by 19 is equal to 221.
Since we don’t know the number being increased by 19, that will be our variable. It will be re-written as so:
N + 19 = 221
Where N is the unknown number.
Using the same method of subtracting by the sum and the number that’s not the variable, we’ll get the solution to N
221-19= 202
N= 202
8.) N + 1 2/3 = 8
8 - 1 2/3 = 19/3= 6 1/3
N= 6 1/3
I really hope this helps
Answer:
13:16 , 52:64
Step-by-step explanation:
The Power of a Product property tells us that when you multiply two numbers with the same base, you can add their exponents to determine the product.
For example,
*
=
.
The Power of a Quotient Property tells us that when you divide two numbers with the same base, you can subtract their exponents to determine the quotient.
For example,
/
=
Answer:
24
Step-by-step explanation:
The question is asking for the net area from x=3 to x=10.
It gives you the net area from x=3 to x=5 being -18.
It gives you the net area from x=5 to x=10, being 42.
Together those intervals make up the interval we want to find the net area for.
-18+42=42-18=24
