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2 years ago

Plzzzzz help will give brainlyist

Mathematics

1 answer:

nikdorinn [45]2 years ago

6 0

Answer:

Step-by-step explanation:

You do 12 x 3 which is 36.

you then do 1/3 x 3 which is 1.

36-1 is 35.

you then do 4x3 which is 12.

you then do 12-5 which is 7.

you then do 35÷7=5

this is why it is 5.

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A kite has vertices at (2, 4), (5, 4), (5, 1), and (0, –1).

gogolik [260]

20 and that should be correct, if you need an explanation just ask

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3 years ago

Help plz I will make u a barinllest

irakobra [83]

A I believe a is correct

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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.

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3 years ago

Ping's Ice Cream Palace offers a special sundae that contains over 2 kilograms of ice cream plus any number of toppings. Write a

hjlf

Let "x" represent the weight of the toppings. We know that we can have any number of toppings. This means that one may ask for no toppings at all too.

Now, we have been told that "S" is the weight of the special sundae in kilograms. This definitely included the "mandatory" 2 kilograms of ice cream. Therefore, S will be at-least equal to 2.

Thus, the inequality that describes S, the weight of the special sundae in kilograms at Ping's Ice Cream Palace is given as:

$S\geq 2+x$ kilograms.

It can be seen that as x increases, S increases too and if an order does not want any toppings in it then the weight of the special sundae will be a minimum of 2 kg which is the weight of the ice cream.

7 0

3 years ago

Read 2 more answers

Given f(x)=-x-1, solve for x when f(x)=-10

Ilia_Sergeevich [38]

f(x) = -10

set the equation equal to -10 and solve for x:

-10 = -x-1

Add 1 to both sides:

-9 = -x

Divide both sides by -1:

x = 9

3 0

3 years ago