Answer: =6xy+x..............
Answer: The answer would be A. Or if that is what you are asking.
Step-by-step explanation:
When you multiply decimals, you multiply like how you would for any other set of numbers. (It doesn't matter if you line the decimal points up or not.)
The only difference between multiplying normal numbers and decimals is the decimal point. How many digits there are to the right of the decimal point is how many times you move the decimal point for the new number to the left.
In the decimals 5.15 and 0.3, there are three digits to the right of the decimal in all, so, when you work this problem and come out with 1545, you move the decimal point to the left three times.
1^545
So the decimal point goes right between the 1 and the 5.
Hope I made it clear enough
Please give me Brainliest
Answer : The length of an edge of cube is, 10 inches.
Step-by-step explanation :
The formula used for area of cube is:
Area of cube = 
where,
a = edge
We are given that:
Area of cube = ![600in^2[tex]Now put all the given values in the above formula, we get:Area of cube = [tex]6a^2](https://tex.z-dn.net/?f=600in%5E2%5Btex%5D%3C%2Fp%3E%3Cp%3ENow%20put%20all%20the%20given%20values%20in%20the%20above%20formula%2C%20we%20get%3A%3C%2Fp%3E%3Cp%3EArea%20of%20cube%20%3D%20%5Btex%5D6a%5E2)
Therefore, the length of an edge of cube is, 10 inches.
Ans is 8 and -1
8 + (-1) = 7
and
8 * (-1) = -8
Answer:
1. y = f(8)
2. So for t which f'(t) > 0, the drug level in the bloodstream is increasing. And for t which f'(t) < 0, it is decreasing.
Step-by-step explanation:
The concentration of the drug in the bloodstream at t hours is:
y = f(t)
1. What is the concentration of the drug in the bloodstream at t= 8 hours?
At t hours, y = f(t)
So at 8 hours, y = f(8)
2. During what time interval is the drug level in the bloodstream increasing? Decreasing?
A function f(t) is increasing when
f'(t) > 0
And is decreasing when
f'(t) < 0
So for t which f'(t) > 0, the drug level in the bloodstream is increasing. And for t which f'(t) < 0, it is decreasing.
