Answer:
137/(36π) ft/s
Step-by-step explanation:
The rate of change of volume will be the product of the rate of change of radius and the area of the sphere. The area of the sphere is ...
A = 4πr² = 4π(6 ft)² = 144π ft²
Then the relationship above is ...
dV/dt = A·dr/dt
548 ft³/s = (144π ft²)·dr/dt
dr/dt = (548 ft³/s)/(144π ft²) = 137/(36π) ft/s ≈ 1.2113 ft/s
Part A:
A coefficient can be either '15' or '25'.
A variable can be either 'w' or 'm'.
A constant is 65.
Part B:
Simply substitute, or plug-in, the numbers and solve.
***Step 1:
65+15w+25m --> 65+15(20)+25(3)
You do this because you are substituting the 'w' for the number
of weeks that Jaxon saved up for, which is 20, and the 'm' for the number
of times that Jaxon mowed the lawn, which is 3.
***Step 2:
65+15(20)+25(3) --> 65+300+75
Begin to solve, using PEMDAS, or whichever acronym you learned.
Remember, if you are using PEMDAS, recall that the order is Parenthesis,
Exponents, Multiplication/Division (whichever comes first), and
Addition/Subtraction (whichever comes first). Here, I checked for parenthesis.
I did find parenthesis, however, they do not have any expressions inside of
them, meaning that these parenthesis are for multiplying, and not for stating
order. So, you skip parenthesis. Next, you check for exponents, which you
find none of, so you skip over that. Now, we get to multiplying/dividing, so
you multiply the 15 and the 20 to get 300, and the 25 and 3 to get 75.
***Step 3:
65+300+75 --> 440
Now, we get to addition. You simply add everything up to get your final
answer: $440.
Part C:
If Jaxon had $75, then yes, the coefficients would change.
By subtracting $65 from $75, we can see that the total amount of money
from Jaxon's deposits and his lawn-mowing money is $10. Jaxon already
deposits $15 a week, meaning that, while using the current equation, Jaxon
CANNOT have $75 in his bank account. We can change the equation
so that Jaxon is able to have $75 in his savings account. You can change
the coefficient of 15 to 10, and the other coefficient of 25 to 0.
Now Jaxon is able to have $75 in his savings account.
16x^4-81y^4 difference of perfect squares...
(4x^2-9y^2)(4x^2+9y^2) that's one of the equivalents...
(2x+3y)(2x-3y)(4x^2+9y^2) that's another one of them...
