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dimulka [17.4K]
3 years ago
7

At the library by 9 books on a certain topic the librarian tells you that 55% of the books on this topic have been signed out ho

w many books does the library have on the topic if 55% of the books have been signed out what percent of the books have not been signed out
Mathematics
1 answer:
Alex17521 [72]3 years ago
7 0
About 5 books have been signed out, so they still have 4.
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A particle is moving on a straight line in such a way that its velocity v is given by v ( t ) = 2 t + 1 for 0 ≤ t ≤ 5 where t is
Lady_Fox [76]

Answer:

The total distance traveled by the particle is S = 30.

Step-by-step explanation:

Given that velocity,

v(t) = 2t + 1

To find the total distance travel, we integrate the velocity function, v(t), to obtain the distance function s(t), and evaluate the resulting distance at the interval given. That is at t = 0 to t = 5.

Integrating v(t) with respect to t, we have

s(t) = t² + t + C.

At t = 5

s(5) = 5² + 5 + C

= 25 + 5 + C

= 30 + C

At t = 0

S(0) = 0 + 0 + C

= C

The required distance is now

S(5) - S(0)

= 30 + C - C

= 30.

8 0
3 years ago
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Find the quotient.<br>8.24 = 1.7=​
WITCHER [35]

Answer:

4.847

Step-by-step explanation:

divide 8.24 by 1.7

8.24 /1.7 = 4.847

6 0
4 years ago
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Need answer asap, correct answer will get brainliest
Cloud [144]

Answer:

x+5

Step-by-step explanation:

When you move to the right or left , that's on the x axis. And since it's to the right, the x- coordinate becomes larger. Hope this helps!

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5. Find the general solution to y'''-y''+4y'-4y = 0
CaHeK987 [17]

For any equation,

a_ny^(n)+\dots+a_1y'+a_0y=0

assume solution of a form, e^{yt}

Which leads to,

(e^{yt})'''-(e^{yt})''+4(e^{yt})'-4e^{yt}=0

Simplify to,

e^{yt}(y^3-y^2+4y-4)=0

Then find solutions,

\underline{y_1=1}, \underline{y_2=2i}, \underline{y_3=-2i}

For non repeated real root y, we have a form of,

y_1=c_1e^t

Following up,

For two non repeated complex roots y_2\neq y_3 where,

y_2=a+bi

and,

y_3=a-bi

the general solution has a form of,

y=e^{at}(c_2\cos(bt)+c_3\sin(bt))

Or in this case,

y=e^0(c_2\cos(2t)+c_3\sin(2t))

Now we just refine and get,

\boxed{y=c_1e^t+c_2\cos(2t)+c_3\sin(2t)}

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4 years ago
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