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masha68 [24]
2 years ago
7

There is a row of 15 trees in a garden. Starting from the first tree, a gardener marked every second tree, then, on his way back

, he marked the first tree, and then every third tree. How many trees are left without a mark?
Mathematics
2 answers:
cluponka [151]2 years ago
7 0
<h3>Answer:  5</h3>

=====================================================

Explanation:

The gardener starts at the first tree and then marks every second tree. So they'll skip a tree each time.

The set of trees he marks has the labels {1,3,5,7,9,11,13,15}. Basically it's the set of odd whole numbers between 1 and 15. Or you could notice that we add on 2 each time we need another tree to mark.

Then he goes back to the first tree and marks every third tree. So we have this new set: {1,4,7,10,13}. Start at 1 and add 3 each time to generate a new item.

Union the two sets mentioned and you'll get {1,3,4,5,7,9,10,11,13,15}

We see that the following trees aren't marked: {2,6,8,12,14}

There are <u>  5  </u> items in that last set.

Side note: The trees with labels {1,7,13} have been marked twice. There's a gap of 6 units between each adjacent label.

zzz [600]2 years ago
4 0

Answer:2

Step-by-step explanation:

15/3 = 5

15/2 = 7.5

+1

= 13

15 - 13 = 2

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Answer:

Therefore rate of payment = $ 3145.72

Therefore the rate of interest = =$1573.17

Step-by-step explanation:

Consider A represent the balance at time t.

A(0)=$ 7864.

r=13 % =0.13

Rate payment = $k

The balance rate increases by interest (product of interest rate and current balance) and payment rate.

\frac{dB}{dt} = rB-k

\Rightarrow \frac{dB}{dt} - rB=-k.......(1)

To solve the equation ,we have to find out the integrating factor.

Here p(t)= the coefficient of B =-r

The integrating factor =e^{\int p(t) dt

                                     =e^{\int (-r)dt

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Multiplying the integrating factor the both sides of equation (1)

e^{-rt}\frac{dB}{dt} -e^{-rt}rB=-ke^{-rt}

\Rightarrow  e^{-rt}dB - e^{-rt}rBdt=-ke^{-rt}dt

Integrating both sides

\Rightarrow \int e^{-rt}dB -\int e^{-rt}rBdt=\int-ke^{-rt}dt

\Rightarrow e^{-rt}B=\frac{-ke^{-rt}}{-r} +C        [ where C arbitrary constant]

\Rightarrow B(t)=\frac{k}{r} +Ce^{rt}

Initial condition B=7864 when t =0

\therefore 7864= \frac{k}{r} - Ce^0

\Rightarrow  C= \frac{k}{r} -7864

Then the general solution is

B(t)=\frac{k}{r}-( \frac{k}{r}-7864)e^{rt}

To determine the payment rate, we have to put the value of B(3), r and t in the general solution.

Here B(3)=0, r=0.13 and t=3

B(3)=0=\frac{k}{0.13}-( \frac{k}{0.13}-7864)e^{0.13\times 3}

\Rightarrow- 0.48\frac{k}{0.13} +11614.98=0

⇒k≈3145.72

Therefore rate of payment = $ 3145.72

Therefore the rate of interest = ${(3145.72×3)-7864}

                                                 =$1573.17

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3 years ago
How to solve this elimination
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