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den301095 [7]
3 years ago
11

20 POINTS FOR ONE SIMPLE QUESTION PLEASE HELP

Mathematics
1 answer:
Sergio [31]3 years ago
7 0
Yeah you did it right but good job!
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What is 1000 divided by 10
Anuta_ua [19.1K]

Answer:

100

Step-by-step explanation:

just take one zero off the end of 1000 to get 100

now if you 100x10 you get 1000

5 0
3 years ago
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Emily and her family are planning to fly to Italy over the summer. If they fly out from LAX, the distance of the trip will be 6,
Iteru [2.4K]

The speed is the ratio between the distance and the time:

s=\frac{d}{t}

Data:

d=6343miles

t=9h

s=\frac{6343\text{miles}}{9h}=704.77\frac{miles}{h}Then, the plane traveled at a speed of 704.77 miles per hour.

8 0
1 year ago
What is equivalent expression 9y+y
andrew11 [14]

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10y

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4 0
3 years ago
Question 1!!! PLEASE HELP
andreev551 [17]
Quickly using a calculator (or long division) reveals that 191/238=<span>0.8025=80.25%, which is close to 80%=4/5
Similarly, 106/160=0.6625=66.25%, which is close to 66.67%=2/3</span>
7 0
3 years ago
HI PLEASE HELP ME WITH MY CALCULUS 1 HW? I AM REALLY STUCK. I need help with parts d,e,g.
asambeis [7]

(d) The particle moves in the positive direction when its velocity has a positive sign. You know the particle is at rest when t=0 and t=3, and because the velocity function is continuous, you need only check the sign of v(t) for values on the intervals (0, 3) and (3, 6).

We have, for instance v(1)\approx-0.91 and v(4)\approx0.91>0, which means the particle is moving the positive direction for 3, or the interval (3, 6).

(e) The total distance traveled is obtained by integrating the absolute value of the velocity function over the given interval:

\displaystyle\int_0^6|v(t)|\,\mathrm dt=\int_0^3-v(t)\,\mathrm dt+\int_3^6v(t)\,\mathrm dt

which follows from the definition of absolute value. In particular, if x is negative, then |x|=-x.

The total distance traveled is then 4 ft.

(g) Acceleration is the rate of change of velocity, so a(t) is the derivative of v(t):

a(t)=v'(t)=-\dfrac{\pi^2}9\cos\left(\dfrac{\pi t}3\right)

Compute the acceleration at t=4 seconds:

a(t)=\dfrac{\pi^2}{18}\dfrac{\rm ft}{\mathrm s^2}

(In case you need to know, for part (i), the particle is speeding up when the acceleration is positive. So this is done the same way as part (d).)

6 0
3 years ago
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