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3 years ago

Show that a∧6 can be written as a perfect square and as a perfect cube.

Mathematics

1 answer:

dezoksy [38]3 years ago

4 0

A^6 = (A^3)^2 = (A^2)^3

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Find the equation of the line that is perpendicular to y = 1/4 x – 2 and passes though the point (5, –2).

zlopas [31]

$\textsf{Let y = ax + b the line we want to find.}$

$\textsf{We know that the slope of y = ax + b is the inverse of the slope of } \mathsf{ y = \frac{1}{4}x - 2.}$

$\textsf{Thus:}$

$\mathsf{a = \dfrac{1}{\frac{1}{4}} = 4}$

$\textsf{As y = 4x + b passes thought (5, -2):}$

$\mathsf{-2 = 4 \cdot 5 + b \Rightarrow 20 + b = -2 \Rightarrow b = -22}$

$\textsf{Hence the line y = 4x - 22 is perpendicular to }\mathsf{y=\frac{1}{4}x - 2}\textsf{ and passes for the point (5, -2).}$

6 0

3 years ago

Read 2 more answers

The number 5/9 is an example of a

Bond [772]

Answer:

5/9 is an example of rational number.

Step-by-step explanation:

as the definition of rational number is the number in the form of p/q but the value of denominator q should not be equal to 0. Every integer is a rational number.

8 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

Helppppppppp please:)

OlgaM077 [116]

Step-by-step explanation:

Thr triangles are similar if any two sides are proportional.

In this, we cannot see any proportional sides, like

DU should be proportional to UT, it is not.

SU should be equal to UE ,it is not.

So, The triangles are not similar.

Hope it helps :)

3 0

3 years ago

There are 92 students in a chemistry class. The instructor must choose two students at random. Students in a Chemistry Class Aca

Sergio039 [100]

Answer:

0.0108 = 1.08% probability that a sophomore non-Chemistry major and then a junior non-Chemistry major are chosen at random.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Probability that a sophomore non-Chemistry major

Out of 92 students, 9 are non-chemistry major sophomores. So

$P(A) = \frac{9}{92}$

Then a junior non-Chemistry major are chosen at random.

Now, there are 91 students(1 has been chosen), of which 10 are non-chemistry major juniors. So

$P(B) = \frac{10}{91}$

What is the probability that a sophomore non-Chemistry major and then a junior non-Chemistry major are chosen at random

$P = P(A)*P(B) = \frac{9}{92}*\frac{10}{91} = \frac{9*10}{92*91} = 0.0108$

0.0108 = 1.08% probability that a sophomore non-Chemistry major and then a junior non-Chemistry major are chosen at random.

8 0

3 years ago