NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by
The sea level is represented by h = 0, therefore, to find the corresponding time when h splashes into the ocean we have to solve for t the following equation:
Using the quadratic formula, the solution for our problem is
The rocket splashes after 26.845 seconds.
The maximum of this function happens at the root of the derivative. Differentiating our function, we have
The root is
Then, the maximum height is
1029.99 meters above sea level.
Answer:
c. g(x) = 4x^2
Step-by-step explanation:
From a first glance, since g(x), is skinnier than f(x), meaning that it is increasing faster, so I know that I can eliminate options A & B since the coefficient on x needs to be greater than 1.
We can then look and see that g(1) = 4 as shown by the point given to us on the graph.
To find the right answer we can find g(1) for options C & D and whichever one matches the point on the graph is our correct answer. e
Option C:
once we plug in 1 for x, our equation looks like
4(1)^2.
1^2 = 1, and 4(1) = 4,
so g(1) = 4. and our point is (1,4).
This is the same as the graph so this is the CORRECT answer.
If you want to double check, you can still find g(1) for option D and verify that it is the WRONG answer.
Option D:
once we plug in 1 for x, our equation looks like
16(1)^2
1^2 = 1, and 16(1) = 16,
so g(1) = 16. and our point is (1,16).
This is different than the graph so this is the WRONG answer.
Answer:
all you have too do is 750*750 = 562,500
Step-by-step explanation:
i hope this helps and pleas give me a stare and a heart thank you so much if you do and if you do not still thank you
Answer:
You can use either of the following to find "a":
- Pythagorean theorem
- Law of Cosines
Step-by-step explanation:
It looks like you have an isosceles trapezoid with one base 12.6 ft and a height of 15 ft.
I find it reasonably convenient to find the length of x using the sine of the 70° angle:
x = (15 ft)/sin(70°)
x ≈ 15.96 ft
That is not what you asked, but this value is sufficiently different from what is marked on your diagram, that I thought it might be helpful.
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Consider the diagram below. The relation between DE and AE can be written as ...
DE/AE = tan(70°)
AE = DE/tan(70°) = DE·tan(20°)
AE = 15·tan(20°) ≈ 5.459554
Then the length EC is ...
EC = AC - AE
EC = 6.3 - DE·tan(20°) ≈ 0.840446
Now, we can find DC using the Pythagorean theorem:
DC² = DE² + EC²
DC = √(15² +0.840446²) ≈ 15.023527
a ≈ 15.02 ft
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You can also make use of the Law of Cosines and the lengths x=AD and AC to find "a". (Do not round intermediate values from calculations.)
DC² = AD² + AC² - 2·AD·AC·cos(A)
a² = x² +6.3² -2·6.3x·cos(70°) ≈ 225.70635
a = √225.70635 ≈ 15.0235 . . . feet
Answer:4(2x + 1) = 12x + 3 - 4x +1=0
Step-by-step explanation:
group like terms =4(2x+1)=12x-4x+3+1
add similar elements=4(2x+1)=8x+3+1
add the numbers=4(2x+1)=8x+4
expand=8x+4=8x+4
subtract 4 from both sides 8x+4-4=8x+4-4
simplify=8x=8x
subtract 8x both sides=0
