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3 years ago

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Damon has five dimes and some nickels in his pocket, worth of a total of $1.20. To find the number of nickels Damon has a, stude

nt wrote the equation 5n + 50 = 1.29. Find the error in the students equation

1 answer:

Agata [3.3K]3 years ago

6 0

The answer to the equation should be in cents whereas it is in dollars which would be incorrect

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For the following system, if you isolated x in the first equation to use the Substitution Method, what expression would you subs

tatiyna

$-x-2y=-4 \\ 3x+y=12 \\ \\ \hbox{the first equation:} \\ -x-2y=-4 \ \ \ \ \ \ |+2y \\ -x=2y-4 \ \ \ \ \ \ \ \ |\times (-1) \\ x=-2y+4$

The expression you would substitute for x is -2y+4. The answer is d.

4 0

4 years ago

Solve for h: -2/3h-4=4/3

Sindrei [870]

Isolate the variable by dividing each side by factors that don't contain the variable.

h = −8

3 0

3 years ago

larisa86 [58]

Answer:

see below

Step-by-step explanation:

f(x) = 5x^3 +1, g(x) = – 2x^2, and h(x) = - 4x^2 – 2x +5

f(-8) = 5(-8)^3 +1 = 5 *(-512) +1 =-2560+1 =-2559

g( -6) = -2 ( -6) ^2 = -2 ( 36) = -72

h(9) = -4( 9)^2 -2(9) +5 = -4 ( 81) -18+5 = -324-18+5=-337

3 0

3 years ago

Read 2 more answers

Which point is on the line that passes through (0,6) and is parallel to the given line? plzz helppp!

Grace [21]

Points on given line = (-12,-2) and (0,-4) because you can see them on the graph. Then draw a parallel line thru (0,6)

To get from (0,-4) to (0,6) your x stays constant and your y coordinate increased by 10. So your new point will do the same in relation to (-12,-2) the x will stay constant at -12 and your y will increase by 10 to +8.

So the answer is A (-12,8)

You can check this because parallel lines have the same slope so

y2-y1/x2-x1 should be equal for both lines.

Line 1: -4 - (-2) / 0 - (-12) = -2/12 = -1/6

Line 2: 6 - 8 / 0 - (-12) = -2/12 = -1/6

5 0

3 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