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3 years ago

This is in my homework paper 5300 divided by 10 to the second

Mathematics

2 answers:

VARVARA [1.3K]3 years ago

6 0

$5300\div 10^2\\5300\div100\\\boxed{53}$

FrozenT [24]3 years ago

4 0

~Take the number 5300 and divide it by 10 squared.
-You should have gotten 5300 divided by 100 (my favorite number because i have no idea why)
~ Next, take 5300 and divide it by 100
-You should have gotten 53.

Sorry that this isn't in all the FANCY typing.

Hope it helped. :)

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vekshin1

Answer:

Step-by-step explanation:

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2 years ago

3 4/7 minus 2 1/5 by rounding to the nearest half

mylen [45]

Answer:

$1 \frac{13}{35}$

Step-by-step explanation:

$3 \frac{4}{7} - 2 \frac{1}{5}$

you can solve for the fractions first, but you have to give them a common denominator

$\frac{20}{35}- \frac{7}{35}$

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then solve for the whole numbers

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3 years ago

At time t ≥ 0, the velocity of a body moving along the s-axis is v = t² -5t +4. When is the body moving backwards

katovenus [111]

Answer:

Step-by-step explanation:

The velocity of a moving body is given by the equation:

$v=t^2-5t+4 ,\, t\geq0$

Is the velocity is positive (v>0), then our object will be moving forwards.

And if the velocity is negative (v<0), then our object will be moving backwards.

We want to find between which interval(s) is the object moving backwards. Hence, the second condition. Therefore:

$v$

By substitution:

$t^2-5t+4$

Solve. To do so, we can first solve for t and then test values. By factoring:

$(t-4)(t-1)=0$

Zero Product Property:

$t=1, \text{ and } t=4$

Now, by testing values for t<1, 1<t<4, and t>4, we see that:

$v(0)=4>0,\, v(2)=-20$

So, the (only) interval for which v is <0 is the second interval: 1<t<4.

Hence, our answer is A.

7 0

3 years ago

Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is

strojnjashka [21]

Answer:

Step-by-step explanation:

The original equation is $121y''+110y'-24y=0$ . We propose that the solution of this equations is of the form $y = Ae^{rt}$ . Then, by replacing the derivatives we get the following

$121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)$

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

$121r^2+110r-24=0$

Recall that the roots of a polynomial of the form $ax^2+bx+c$ are given by the formula

$x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}$

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

$r_1 = -\frac{12}{11}$

$r_2 = \frac{2}{11}$

So, in this case, the general solution is $y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}$

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

$c_1 + c_2 = 1$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 0$ (or equivalently $c_2 = 6c_1$

By replacing the second equation in the first one, we get $7c_1 = 1$ which implies that $c_1 = \frac{1}{7}, c_2 = \frac{6}{7}$ .

So $y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}$

b) By using y(0) =0 and y'(0)=1 we get the equations

$c_1+c_2 =0$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 1$ (or equivalently $-12c_1+2c_2 = 11$

By solving this system, the solution is $c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}$

Then $y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}$

The Wronskian of the solutions is calculated as the determinant of the following matrix

$\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2$

By plugging the values of $y_1$ and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

$e^{\int -p(x) dx}$

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

$e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}$

Note that this function is always positive, and thus, never zero. So $y_1, y_2$ is a fundamental set of solutions.

8 0

3 years ago

If our monthly income is $3050 in your house payment is $2400 what fraction of your monthly income must go to pay your house pay

ivolga24 [154]

Answer:

The 28% rule states that you should spend 28% or less of your monthly gross income on your mortgage payment (e.g. principal, interest, taxes and insurance). To determine how much you can afford using this rule, multiply your monthly gross income by 28%

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3 years ago