The population of a large U.S. city is2,707,210. Which of the following best expresses this population?

Kevin is 4 times as old as Daniel and is also 6 years older than Daniel

MAXImum [283]

Answer:

Part a) Daniel's age is 2 years

Part b) Kevin's age is 8 years

Step-by-step explanation:

The question is

Part a) How old is Daniel?

Part b) How old is Kevin?

Let

x ----> Kevin's age

y ----> Daniel's age

we know that

$x=4y$ -----> equation A

$x=y+6$ ----> equation B

Equate equation A and equation B

$4y=y+6$

solve for y

$4y-y=6$

$3y=6$

$y=2$

therefore

Daniel's age is 2 years

Find the value of x

substitute the value of y in any of the two equations

$x=4(2)=8\ years$

$x=2+6=8\ years$

therefore

Kevin's age is 8 years

6 0

2 years ago

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

YOOO MARKIN BRAINIEST !!

Brilliant_brown [7]

Y ≥ 2x-5
y-intercept is -5, from there go 2 up 1 over ( slope is rise over run)
greater than or equal so the line is connected
now find the shaded area by plugging in (0,0)
0 ≥2(0)-5
0 ≥-5 is correct so shade the area that to the left of the line, (the whole area that including (0,0))

y<-3x
y intercept is 0 so start from there and go down 3 right 1 (or go up 3 left 1)
broken like cause no or equal sign
the (0,0) is on the line so use (1,1) to find the answer
1<-3(1) is incorrect so shade the area that dies not include (1,1) or the entire area to the left of that line

you can see the section where both shaded area cross, thats the answer so erase every area you shaded that isn’t the answer so
THE ANSWER IS C

3 0

3 years ago

Read 2 more answers

in january, meghan made 12 fewer sales calls than paul did. together they made 256 calls. how many calls did meghan make that mo

adelina 88 [10]

2÷256=128 then subtract 128-12=116

3 0

3 years ago

Write the equation of a quadratic function in standard form for the parabola that passes through the given points.

vazorg [7]

Answer:

5,7 given point its korrect now

8 0

1 year ago