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3 years ago

7

- 7th grade work - Please help, and answer only if you have an answer.

2 answers:

Schach [20]3 years ago

5 0

$268.38 is the correct answer

masha68 [24]3 years ago

4 0

Answer:

$268.38

Step-by-step explanation:

First we will take off the 20% off the 315 dollars. SO we can do this by saying that the discounted price is 80% of the original price, (because 100 - 20 = 80). So 0.8 * 315 = 252. So 252 = discounted price. So then we have to add the tax price.

Sales tax is %6.5, so we can do 252*0.065 to find the tax payed. We get 16.38, so we can dd this to our discounted price to get 252 + 16.38, or $268.38

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Jamal ran 7 kilometers, sprinted 400 meters, and climbed bleachers for a distance of 1.6 kilometers during track practice. How m

seraphim [82]

First, convert 7 kilometers to meters, which is 7000 meters.
Add that to 400, and now you he ran 7400 meters.
Now convert 1.6 kilometers to meters, which is 1600 meters.
Add that to 7400, and the total is 9000 meters.

Hope this helped!

4 0

3 years ago

Read 2 more answers

The chef has 80 pounds of strip loin. Th

PIT_PIT [208]

Um i was thinking 5% or something like that

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3 years ago

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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

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prohojiy [21]

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3 years ago

Type the simple subject of this sentence.the plane leaves at 6:16 and arrives at 10:50.

Jet001 [13]

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3 years ago