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Mkey [24]
3 years ago
9

The population of bees has been decreasing at a rate of 10% per year in 2012 there were 6000 bees in particular hive how many be

es were there in 2014
Mathematics
1 answer:
natulia [17]3 years ago
6 0
2012 = 6000 bees
10% of  6000 is 60
so 6000 - 60 = 5940
2013 = 5940 bees
 10% of 5940 = 59.4
so 5940 - 59.4 = 5880.6
so in 2014 you 5881 because you cant have a 0.6 of a bee
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yKpoI14uk [10]

Answer:

THESE CAN HELP U OUT

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B=15

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2 years ago
I need help please is this question i needed ;-;. What amount in dollars would Peter save on a $64 coat that is on sale for 30 o
Alchen [17]

Answer:

He would save $19.20.

Step-by-step explanation:

30% = 0.3

64*0.3 = 19.2

There he saves $19.20 on the coat.

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2 years ago
Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de
Ganezh [65]

Answer:

a. \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

b. \mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

Step-by-step explanation:

The initial value problem is given as:

y' +y = 7+\delta (t-3) \\ \\ y(0)=0

Applying  laplace transformation on the expression y' +y = 7+\delta (t-3)

to get  L[{y+y'} ]= L[{7 + \delta (t-3)}]

l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

Taking inverse of Laplace transformation

y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{1}{s+1}] = e^{-t}  = f(t) \ then \ by \ second \ shifting \ theorem;

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

= e^{-t-3} \left \{ {{1 \ \ \ \ \  t>3} \atop {0 \ \ \ \ \  t

= e^{-(t-3)} u (t-3)

Recall that:

y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

Then

y(t) = 7 -7e^{-t}  +e^{-(t-3)} u (t-3)

y(t) = 7 -7e^{-t}  +e^{-t} e^{-3} u (t-3)

\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

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2 years ago
How many vertices does a heptagon have
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Hey! There are 7 sides and 7 vertices on a Heptagon. Hope this helps!
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3 years ago
Read 2 more answers
Two angles of a triangle have the same measure and the third one is 9 degrees greater than the measure of each of the other two.
Umnica [9.8K]
The largest angle should be 66°
I hope this helps.
YOU'RE WELCOME :D
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