What is the estimate for 200,629+28,542

2 answers:

5 0

If you want it to be easier to add, just do it like this: 200,000+ 30,000= 230,000
If you want me to be more specific, put this: 201,000+ 29,000= 230,000
It really doesn't matter because both estimates are the same answer, so you can put whichever one. Hope this helps!!!! :)

vladimir2022 [97]2 years ago

4 0

Answer:

The estimate of 200,629+28,542 is 230000.

Step-by-step explanation:

Given : Expression 200,629+28,542

To find : What is the estimate for expression ?

Solution :

Estimate the number 200,629 by rounding we get approximately

$200629\approx 200000$

Estimate the number 28,542 by rounding we get approximately

$28542\approx 30000$

So, If we add these number we get

$200000+30000=230000$

Therefore, The estimate of 200,629+28,542 is 230000.

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A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$