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Katarina [22]
2 years ago
6

Help me i will give brainlist

Mathematics
2 answers:
taurus [48]2 years ago
7 0

Answer:

1)=87.71 2)=98.37 3)=4836.9 4)=310.76 5)=405.856

Step-by-step explanation:

if u need more contact me

elena55 [62]2 years ago
3 0

Answer: heres the answer

Step-by-step explanation:

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Is 1600 oz greater than or less than 10 pounds
kipiarov [429]
It's greater then......
5 0
3 years ago
Read 2 more answers
We are standing on the top of a 1680 ft tall building and throw a small object upwards. At every second, we measure the distance
MAVERICK [17]

Answer:

a) The height of the small object 3 seconds after being launched is 2304 feet.

b) The small object ascends 128 feet between 5 seconds and 7 seconds.

c) The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) The object will take 21 seconds to hit the ground.

Step-by-step explanation:

The correct formula for the height of the small object is:

h(t) = -16\cdot t^{2}+256\cdot t+1680 (1)

Where:

h - Height above the ground, measured in feet.

t - Time, measured in seconds.

a) The height of the small object at given time is found by evaluating the function:

h(3\,s)= -16\cdot (3\,s)^{2}+256\cdot (3\,s)+1680

h(3\,s) = 2304\,ft

The height of the small object 3 seconds after being launched is 2304 feet.

b) First, we evaluate the function at t = 5\,s and t = 7\,s:

h(5\,s)= -16\cdot (5\,s)^{2}+256\cdot (5\,s)+1680

h(5\,s) = 2560\,s

h(7\,s)= -16\cdot (7\,s)^{2}+256\cdot (7\,s)+1680

h(7\,s) = 2688\,s

We notice that the small object ascends in the given interval.

\Delta h = h(7\,s)-h(5\,s)

\Delta h = 128\,ft

The small object ascends 128 feet between 5 seconds and 7 seconds.

c) If we know that h = 2640\,ft, then (1) is reduced into this second-order polynomial:

-16\cdot t^{2}+256\cdot t-960=0 (2)

All roots of the resulting equation come from the Quadratic Formula:

t_{1} = 10\,s and t_{2}= 6\,s

The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) If we know that h = 0\,ft, then (1) is reduced into this second-order polynomial:

-16\cdot t^{2}+256\cdot t +1680 = 0 (3)

All roots of the resulting equation come from the Quadratic Formula:

t_{1} = 21\,s and t_{2} = -5\,s

Just the first root offers a solution that is physically reasonable.

The object will take 21 seconds to hit the ground.

8 0
2 years ago
v = 20.8x^2 - 458.3x + 3,500 represents the value of a car from 1964 to 2002. What year did the car have the least value? (x= 0
Karo-lina-s [1.5K]
See the picture for all detail, I have done it by hand.
Few important steps;
-Take the first derivative
-Put it equal to zero
-you get the x value which is equal to 11
-Now for the check that x=11 is the least value or not, take the second derivative
-As, the 2nd derivative is positive so it is the least value

As x=0 in 1964 add 11 to it so at x=11 we get 1975 so in this year car have least value

7 0
3 years ago
How to solve that problem ?
dangina [55]
It a really hard problem for a 5th grader

4 0
3 years ago
The perimeter of the square below can be described by the expression 64 + 12y.
Fantom [35]

Answer:

the answer would be 16+3y

Step-by-step explanation:

length of one side of a square is perimeter/4

so (64+12y)/4

64/4 +12y/4

=16+3y

8 0
2 years ago
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