Answer:
156 in squared
Step-by-step explanation:
So the formulas will be stated below
Parallelogram Area= 1/2h(b1+b2)
Square Area= Bh
So first do
(4)(12+6)
(4)(18)
No need to half it bc there's the same shape at the bottom
72+18+30+36
6(3)
3x5(2)
12(3)
Answer:
Raising something to a negative exponent is just taking the reciprocal of the amount.
Step-by-step explanation:
Let's assume that you wanted to know what
is.
To find it, you would take the reciprocal of the x amount. So
becomes
.
This works because of the nature of exponents. Exponents represent the number of times you are multiplying a value by itself. So
would be equal to a · a · a. To increase the exponent, you increase the number of times the value is multiplied by itself: To increase
to
, you would have to multiply a with
two more times (a · a · a · a · a). To decrease the exponent, you must divide the value by itself. So to decrease
to
, you would have to divide
by a 3 times.
If the exponent is 0, the value is equal to 1. But you can still decrease the exponent into negative numbers. You just divide 1 by a the desired amount of times:
means that you are dividing 1 by a 3 times.
Hope this helps.
Answers 8 football, 6 baseball and basketball, 5 soccer
Step-by-step explanation:
8 football, 6 baseball and basketball, 5 soccer
Answer:

Step-by-step explanation:


taking like terms together

taking LCM


taking LCM

splitting the term

splitting the term



we know that

putting this value in above equation
