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lesantik [10]
3 years ago
13

I need help on this?

Mathematics
1 answer:
vampirchik [111]3 years ago
7 0
Answer:

it’s 65

Explanation:

because I went within
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Which statement is true about the points and planes?
Reil [10]

Answer:

The answer is B

Step-by-step explanation:

Ed2020

3 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
Zhane picked 28 cherries. 1/7 of them are bad. How many groups of cherries would she have to divide them into to find out how ma
castortr0y [4]
She would have 7 groups of cherries
4 0
3 years ago
If triple the difference of a number and two is three more than the number.
aksik [14]

Answer:

Rephrase that so that x=the number.

3+2x=x

(-2x) => 3=-x

(*-1) => -3=x

Therefore the number is -3!

Step-by-step explanation:

hope this helps :)

5 0
2 years ago
(2x^2+4x-3)-(2x^2+4x-3) show work mark brainliest
Aleksandr [31]

Solution: \left(2x^2+4x-3\right)-\left(2x^2+4x-3\right)=0

Steps:

\mathrm{Remove\:parentheses}:\quad \left(a\right)=a, =2x^2+4x-3-\left(2x^2+4x-3\right)

-\left(2x^2+4x-3\right):\quad -2x^2-4x+3

=2x^2+4x-3-2x^2-4x+3

\mathrm{Simplify}\:2x^2+4x-3-2x^2-4x+3:\quad 0

=0

8 0
3 years ago
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