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3 years ago

Identify the percent, amount, and base in this problem. What is 90% of 70?

1 answer:

4 0

The answer here is D, Percent = 90%, Amount = unknown, Base = 70

We can solve this by using what we know. The first step is to note that 90 is the percent (since it has the percent sign). In addition, we know that 70 is the base (since we're taking a percent of 70). That gives us our answer of D, Percent = 90%, Amount = unknown, Base = 70.

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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

2 years ago

Write a statistical question to find out students favorite flavor of gum

igomit [66]

25% of the class likes mint gum, 5% likes orange, 20% likes raspberry lemonade, of the remaining % likes cherry. how much is the remaining % of students?

6 0

3 years ago

Read 2 more answers

a person in a whispering gallery standing at one focus of the ellipse can whisper and be heard by a person standing at the other

Lynna [10]

The Height of the ceiling at the center is 57.6628 feet

In geometry, an ellipse is a two-dimensional shape, that is defined along its axes. An ellipse is formed when a cone is intersected by a plane at an angle with respect to its base.

It has two focal points. The sum of the two distances to the focal point, for all the points in curve, is always constant.

The General form of ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

According to the question,

Length of whispering gallery is 130 feet

The foci are located 30feet from the center (0,0)

So, 2a = 130

=> a = 130/2

=> a = 65

and c = 30

As we know ,

=> c² = a² - b²

=> (30)² = (65)² - b²

=> 900 = 4225 - b²

=> b² = 4225 - 900

=> b² = 3325

=> b = 57.7 feet

Therefore , the height of the ceiling at the center is 57.6628 feet

To know more about Ellipse here

brainly.com/question/19507943

#SPJ4

7 0

10 months ago

roger stan and steve have a standard deck of cards each of them draw a card from the deck find the probability of roger drawing

tatiyna

roger

1/5

steve

0/5

stan

1/5

5 0

3 years ago

The quantity a varies directly with b and c and inversely with dThe quantity d is tripledWhich of the following must be true for

EastWind [94]

Answer:

We have the next relation:

A = (b*d)/c

because we have direct variation with b and d, but inversely variation with c.

Now, if we have 3d instead of d, we have:

A' = (b*(3d))/c

now, we want A' = A. If b,c, and d are the same in both equations, we have that:

3bd/c = b*d/c

this will only be true if b or/and d are equal to 0.

If d remains unchanged, and we can play with the other two variables we have:

3b'd/c' = bd/c

3b'/c' = b/c

from this we can took that: if c' = c, then b' = b/3, and if b = b', then c' = 3c.

Of course, there are other infinitely large possible combinations that are also a solution for this problem where neither b' = b or c' = c

6 0

3 years ago