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stiv31 [10]
3 years ago
8

Here are some brain points comment on this if u want more and pls don't report thats just mean

Mathematics
2 answers:
tamaranim1 [39]3 years ago
7 0

Answer:

Step-by-step explanation:

SashulF [63]3 years ago
5 0

Answer:

thanks

Step-by-step explanation:

You might be interested in
An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object
Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2  + \frac{1}{2} at^2\\3) v = v_0 + at

What do we already know?

  • v_0 = 29.4 ms^-1
  • The launch is straight up
  • a = -9.81 ms^-2 this is the gravitational acceleration g
  • s_0 = 0 m, since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, v = 0 ms^-2

v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2

this is the equation of the whole trajectory that object makes.

and if you solve this by making s = 0, you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)

5 0
3 years ago
A fruit market equally divided 80 red apples into 8 containers, and equally divided 54 green apples into 6 containers. Natalie b
Goryan [66]

Answer:

10 + 9 = n

19 apples

Step-by-step explanation:

80 ÷ 8 = y and 56 ÷ 6 = a

n = y + a

80 ÷ 8 × 8 = y × 8

80 = 8y

80 ÷ 8 = 8y ÷ 8

y = 10

54 ÷ 6 = a

54 ÷ 6 × 6 = a × 6

6a = 54

6a ÷ 6 = 54 ÷ 6

a = 9

10 + 9 = n

19 = n

7 0
3 years ago
Which is greater 9/20 or 60%
Readme [11.4K]

Answer:

60%

Step-by-step explanation:

9/20 is 45%

8 0
4 years ago
Read 2 more answers
Please a genius help me please
Vedmedyk [2.9K]

Answer:

B, D, and F

They match the graph

6 0
2 years ago
Write a rule for the nth term of the arithmetic sequence:<br><br> a11= 50, d = 7
ivolga24 [154]

Answer:

Required rule for n^{th} is a_{n}=7n-27.

Step-by-step explanation:

Given that,

a_{11} = 50,\  \  d=7

From the question: we have to write the n^{th} term of Arithmetic sequence.

Arithmetic Sequence or Arithmetic progression (A.P) : It is a sequence which possess that difference between of two successive sequence is always constant.

                a_{1} ,a_{2},a_{3},a_{4}.....................a_{n-1},a_{n}

                                        where, a_{1} is the first term of A.P

                                                     d is the common difference.

                                                     a_{n} is the last term or general term.

The above sequence to be in A.P then their common difference should be equal.

          d=a_{2}-a_{1} =a_{3}-a_{2}=a_{4} -a_{3} ..........................a_{n}-a_{n-1}

Now, Formula of General Term is a_{n}=a+(n-1)d

So,                                                    a_{11}= a+(11-1)d\\        a_{11} = a+10d

 Substituting the value of   a_{11} = 50,\  \  d=7 we get,

                                                         50=a+10\times7\\50=a+70\\a=-20

Then General term (a_{n}) of given data is

                                                         a_{n}=-20+(n-1)7\\a_{n}=-20+7n-7\\a_{n}=7n-27

Therefore, Required rule for n^{th} is a_{n}=7n-27.

5 0
4 years ago
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