Ordering numbers from greatest to least is called putting them in ____

Evaluate <br> 3/4•1/2<br><br> A.3/8<br> B.4/6<br> C.2/12<br> D.1/6

Kisachek [45]

Answer: (A) 3/8

Step-by-step explanation:

3/4*1/2

3/4*2

= 3/8

4 0

3 years ago

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Q1 A ball is thrown upwards with some initial speed. It goes up to a height of 19.6m and then returns. Find (a) The initial spee

lubasha [3.4K]

Answer:

(a) 19.6 ms⁻¹

(b) 2 s

(c) 9.8 ms⁻¹

(d) 4 s

Step-by-step explanation:

<u>Constant Acceleration Equations (SUVAT)</u>

$\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}$

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

Acceleration due to gravity = 9.8 ms⁻².

When the ball reaches its maximum height, its velocity will momentarily be zero.

<u>Given values</u> (taking up as positive):

$s=19.6 \quad v=0 \quad a=-9.8$

$\begin{aligned}\textsf{Using} \quad v^2&=u^2+2as\\\\\textsf{Substitute the given values:}\\0^2&=u^2+2(-9.8)(19.6)\\0&=u^2-384.16\\u^2&=384.16\\u&=\sqrt{384.16}\\\implies u&=19.6\; \sf ms^{-1}\end{aligned}$

Therefore, the initial speed is 19.6 ms⁻¹.

Using the same values as for part (a):

$\begin{aligned}\textsf{Using} \quad s&=vt-\dfrac{1}{2}at^2\\\\\textsf{Substitute the given values:}\\19.6&=0(t)-\dfrac{1}{2}(-9.8)t^2\\19.6&=4.9t^2\\t^2&=\dfrac{19.6}{4.9}\\t^2&=4\\t&=\sqrt{4}\\\implies t&=2\; \sf s\end{aligned}$

Therefore, the time taken to reach the highest point is 2 seconds.

As the ball reaches its maximum height at 2 seconds, one second before this time is 1 s.

<u>Given values</u> (taking up as positive):

$u=19.6 \quad a=-9.8 \quad t=1$

$\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=19.6+(-9.8)(1)\\v&=19.6-9.8\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}$

The velocity of the ball one second before it reaches its maximum height is the <u>same</u> as the velocity one second after.

<u>Proof</u>

When the ball reaches its maximum height, its velocity is zero.

Therefore, the values for the downwards journey (from when it reaches its maximum height):

$u=0 \quad a=9.8 \quad t=1$

(acceleration is now positive as we are taking ↓ as positive).

$\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=0+9.8(1)\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}$

Therefore, the velocity of the ball one second before <u>and</u> one second after it reaches the maximum height is 9.8 ms⁻¹.

From part (a) we know that the time taken to reach the highest point is 2 seconds. Therefore, the time taken by the ball to travel from the highest point to its original position will also be 2 seconds.

Therefore, the total time taken by the ball to return to its original position after it is thrown upwards is 4 seconds.

4 0

1 year ago

Anybody need any 4rth 5th 3rd grade math help or reading or writing im here!!

Klio2033 [76]

Answer:

yeah me too just mark me brainliest so I can get points plllzzzz

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Find each sum. (-3)+2

iren2701 [21]

Hope this will help

4 0

3 years ago

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Admission to an amusement park is $10 for children and $15 for adults.On saturday 1,300 people enter the park and $17,250 is col

Natali [406]

We have to make 2 equations
A = Adults C= Children
So: A + C = 1300
15A + 10C = 17250
We can solve for A and C in 2 ways, SUBSTITUTION or ELIMINATION
I'll do substitution
SUBSTITUTION: Subtract A from both sides
A + C = 1300
-A -A
C = 1300 - A
Plug C is to the 2nd equation
15A + 10(1300 - A) = 17,250
15A + 13,000 - 10A = 17,250
5A + 13,000 = 17,250
- 13,000 -13,000
5A = 4250
---- ---------
5 5
A= 850
So, there are 850 adults, subtract 850 from 1,300 and there 450 children
Answer: 850 adults
450 children

3 0

3 years ago

Ordering numbers from greatest to least is called putting them in _____ order.