Weird way to write it but alright! (Sideways)
19pq^-2 x 5pq^6 = ?
These problems are pretty much single operations between each of the variables / constants.
So it's like this:
(19*5)(p*p)(q^-2*q^6) = ?
19*5 is 95.
For p*p remember that when two variables multiply there given powers add. In the case where the powers are not shown (like in the case of p*p) they are always assumed to be 1. So what is 1+1? 2.
p*p is p^2
For q^-2*q^6 it is the same deal with the previous problem. So now the problem looks like this:
-2 + 6 = 4
(The two is negative, because the power is negative 2)
So, q^4.
Our final answer is all of the combined.... like a so:
95p^2q^4
Answer:
11 + Mai's Score = 59
Step-by-step explanation:
You need to add 11 and Mai's score together to get 59, so with the values given we can make the equation 11 + Mai's Score = 59.
*depending on the question, Mai's score may need to be said as a letter variable, so:
If m = mai's score,
11 + m = 59
I hope this helped! :)
9514 1404 393
Answer:
$7.14
Step-by-step explanation:
Let p, d, q represent the numbers of pennies, dimes, and quarters in the collection, respectively.
p + d + q = 45 . . . . . . . . there are 45 coins in the collection
2p +5 = q . . . . . . . . . . . . 5 more than twice the number of pennies
p + 4 = d . . . . . . . . . . . . . 4 more than the number of pennies
Substituting the last two equations into the first gives ...
p +(p +4) +(2p +5) = 45
4p = 36 . . . . . . . . . . . . . subtract 9
p = 9 . . . . . . . . . . . divide by 4
d = 9 +4 = 13
q = 2(9) +5 = 23
The value of the collection is ...
23(0.25) +13(0.10) +9(0.01) = 5.75 +1.30 +0.09 = 7.14
The coin collection is worth $7.14.
It’s line 4. She subtracted the final from the initial. The distance formula requires you to subtract the initial from the final. That’s the formula for displacement in x and y.
Answer:
-17.8884
Step-by-step explanation:
- 6.444 - 3.2222 - 8.2222 = -17.8884
