Answer:
C
Step-by-step explanation:
So we have the function:
![h(t)=120t-16t^2](https://tex.z-dn.net/?f=h%28t%29%3D120t-16t%5E2)
Where h(t) measures the height after t seconds.
To find the height of the object after 6 seconds, substitute 6 for t. So:
![h(t)=120(6)-16(6)^2](https://tex.z-dn.net/?f=h%28t%29%3D120%286%29-16%286%29%5E2)
Evaluate. Square:
![h(t)=120(6)-16(36)](https://tex.z-dn.net/?f=h%28t%29%3D120%286%29-16%2836%29)
Multiply:
![h(6)=720-576](https://tex.z-dn.net/?f=h%286%29%3D720-576)
Subtract:
![h(6)=144](https://tex.z-dn.net/?f=h%286%29%3D144)
So, the height after six seconds of the object is 144 ft.
Our answer is C.
Least common multiple of 60 is code for "it's a really small number."
Less than or equal to 12 confirms that theory.
GCF of the two numbers is 2 tells me "it's the smallest numbers around."
The answer is 2 and 4. Their multiples can arrive at 60, they are less than 12, and the GCF for both numbers is 2.
Answer:
true
Step-by-step explanation:
i think
Answer: I would say A because it makes the most sense
Step-by-step explanation:
Let Fn be the number of ways of arranging such flagpole with the given conditions.
When arranging a flagpole of n feet high, consider the following cases
If the last flag used is a red flag, then the other flags are n-1 foot high, so they can be seen as arranged on a smaller flagpole of n-1 feet high, which can be done in Fn-1 ways.
Similarly, If the last flag used is a gold flag, then the other flags can be seen as arranged on a smaller flagpole of n-1 feet high. This can be done in Fn-1 ways.
If the last flag used is green, the other flags are n-2 feet high, so the flagpole can be arranged in Fn-2 ways.
Using the sum rule, we obtain that Fn = 2Fn-1 + Fn-2 for all n≥3. Listing all the combinations of flags, the initial conditions are F1 = 2 and F2 = 3.
To Learn more about similar flagpole questions:
brainly.com/question/14277535
#SPJ4