Answer:
26
Step-by-step explanation:
[(7+3)5-4]/2+3
-To solve this equation you have to use PEMDAS
P- Parentheses
E- Exponents
M- Multiplication
D- Division
A- Addition
S- Subtraction-
- With MD and AS you work left to right of the equation since they are in the same spot. (PE[MD][AS])
Step 1) [(10)5-4]/2+3
- First you do "P," parentheses, so you add 7+3=10
Step 2) [50-4]/2+3
- Next you do "M," multiplication, and multiply 10x5=50
Step 3) [46]/2+3
- Then you do "S," subtraction, and subtract 50-4=46
(FYI: Steps 1-3 were still in the parentheses. We had to start with the parentheses in the parentheses, work PEMDAS, and now we are out of the parentheses and have to work PEMDAS on the rest of the problem.)
Step 4) 23+3
- Now we do "D," division, and divide 46/2=23
Step 5) 23+3=6
- Finally we do "A," addition, and add 23+3=26 so the answer is 26
(FYI: "/" means division)
D ~ Positive Exponetial growth. Hope it helped!!! <3 (^u^)
Answer:
12
Step-by-step explanation:
This can be solved by working backwards.
7 is one more than half the number of invitations.
Subtract 1. 6 is half the number of invitations.
Double.
12 is the full number of invitations.
Algebra (if you must!):
x = number of invitations
x/2 + 1 = 7
Subtract 1.
x/2 = 6
Multiply by 2.
x = 12
One possible system is
1x + 3y = 4
2x + 6y = 8
Note how 2 is twice as large as 1, 6 is twice as large as 3, and 8 is twice as large as 4.
In other words, the second equation is the result of multiplying both sides of the first equation by 2.
1x+3y = 4
2*(1x+3y) = 2*4
2x+6y = 8
Effectively the two equations in bold are the same which produces the same line. The two lines overlap perfectly to intersect infinitely many times. An intersection is a solution.
