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3 years ago

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.

_____

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.

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

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:

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

3 0

3 years ago

Write an equation of the line below.

otez555 [7]

Answer:

im not exactly sure but i think it would be y=1x-2

7 0

2 years ago

10q - 30 is equivalent to _____.

mr_godi [17]

For this case we must find an expression equivalent to:

$10q-30$

We look for the factors of 10 and 30:

10: 1,2,5,10

30: 1,2,3,5,6,10,15,30

The greatest common factor of both numbers is 10.

Then, we can rewrite the expression as:

$10 (q-3)$

Answer:

$10 (q-3)$

5 0

3 years ago