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3 years ago

All of these methods can be used to show that two triangles are similar EXCEPT

Mathematics

2 answers:

irga5000 [103]3 years ago

8 0

the correct answer 'SS'

kvasek [131]3 years ago

7 0

The answer is probably ss or SSA depending on your answer choices, if one of these isnt an answer choice let me know

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What is the slope of the line 2x+5y=25

natali 33 [55]

You can put this solution on YOUR website! You must first get the equation into slope-intercept format: y = mx + b. . 2x+-+5y+=+-25 . Subtract 2x from both sides

6 0

3 years ago

Read 2 more answers

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

what is the slope intercept form of the linear equation with a graph that passes through (-2,2) and is perpendicular to the grap

qwelly [4]

Hey there!!

Steps to get into equation which is perpendicular:

Change the slope

The slope of a perpendicular line is always negative reciprocal.

Take the point and get that into point-slope form.

y - y = m ( x - x )

Convert this equation into slope-intercept form.

'm' is the slope.

We have the slope as 1/3

The perpendicular slope would be -3.

Point slope:

The point = (-2,2)

y - 2 = -3( x - (-2) )

y - 2 = -3(x+2)

y - 2 = -3x -6

Adding 2 on both sides:

y = -3x - 4

Hence, the equation would be :

y = -3x - 4

Hope it helps!

7 0

3 years ago

In a recent golf match, Tiger’s score was 4 less than Phil’s score. Their combined scores totaled 140. Let p represent Phil's sc

Svetach [21]

The answer would be D .

3 0

3 years ago

5. What is the slope of the line ?

guapka [62]

Answer:

it is 2/3

Step-by-step explanation:

6 0

3 years ago