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andrey2020 [161]
3 years ago
13

The midpoint of (-4,15) and (22,3)

Mathematics
1 answer:
ahrayia [7]3 years ago
8 0

Answer:

(9, 9)

Step-by-step explanation:

midpoint of (-4,15) and (22,3)

=  \bigg( \frac{ - 4 + 22}{2}, \:  \:  \frac{15 + 3}{2}  \bigg) \\  \\  =  \bigg( \frac{18}{2},  \:  \:  \frac{18}{2}  \bigg) \\  \\  =  ( 9,  \:  \:  9) \\  \\

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Isabella save 2 nickels today, if she doubles the number of Nicole, she saves every day, how many days, including today, will it
Elden [556K]

Answer:

  8 days

Step-by-step explanation:

On day 8, Isabella will save 256 nickels, bringing her total to 510.

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The number of nickels saved on day n is 2^n. The total is 2^(n+1)-2.

_____

The above can be written down from your knowledge of binary sequences. If you want a more formal development, read on.

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The number of nickels saved on day n is a geometric sequence with first term 2 and common ratio 2. The n-th term of the sequence is ...

  an = a1·r^(n-1) = 2·2^(n-1) = 2^n

The sum of n terms of the sequence is ...

  S = a1(r^n -1)/(r -1) = 2(2^n -1)/(2-1)

  S = 2^(n+1) -2

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We want S > 500, so ...

  500 < 2^(n+1) -2

  502 < 2^(n+1)

  251 < 2^n

  log(251) < n·log(2)

  n > log(251)/log(2)

  n > 7.97 . . . . . . . . 8 days or more to save more than 500 nickels

3 0
3 years ago
Jess walked from her home to her friend’s house and arrived at 4.15. If it took Jess 25 minutes to walk this distance, then what
Airida [17]
D.3:50

Count back 25 minutes from 4:15 and you’ll get 3:50. The numbers on the clock are 5 minutes apart. It should be easy to solve just by using a standard clock
3 0
2 years ago
Read 2 more answers
x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the seco
igomit [66]

Answer:

x=-cos(t)+2sin(t)

Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0

Will have a characteristic equation of the form:

a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0

Where solutions r_1,r_2...,r_n are the roots from which the general solution can be found.

For real roots the solution is given by:

y(t)=c_1e^{r_1t} +c_2e^{r_2t}

For real repeated roots the solution is given by:

y(t)=c_1e^{rt} +c_2te^{rt}

For complex roots the solution is given by:

y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)

Where:

r_1_,_2=\lambda \pm \mu i

Let's find the solution for x''+x=0 using the previous information:

The characteristic equation is:

r^{2} +1=0

So, the roots are given by:

r_1_,_2=0\pm \sqrt{-1} =\pm i

Therefore, the solution is:

x(t)=c_1cos(t)+c_2sin(t)

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of x(t) in order to find the constants c_1 and c_2:

x'(t)=-c_1sin(t)+c_2cos(t)

Evaluating the initial conditions:

x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1

And

x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2

Now we have found the value of the constants, the solution of the second-order IVP is:

x=-cos(t)+2sin(t)

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3 years ago
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otez555 [7]

Answer:

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7 0
2 years ago
10q - 30 is equivalent to _____.
mr_godi [17]

For this case we must find an expression equivalent to:

10q-30

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Then, we can rewrite the expression as:

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Answer:

10 (q-3)

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