How do solve this problem? 1 2/5x =7

A right triangle has a hypotenuse of length 99 inches. if one angle is 34 degrees34°, find the length of each leg.

Andreyy89

Sine (34) = opposite / 99
opposite side = .55919 * 99
opposite side = 55.35981 

Adjacent side² = 99² - 55.35981²
Adjacent side² = 9,801 -3,064.7085632361 
Adjacent side² = 6,736.29 
Adjacent side = 82.0749

5 0

2 years ago

Quick calculus question work doesn't have to be shown

Vinil7 [7]

$\bf \textit{volume of a sphere}\\\\
V=\cfrac{4\pi r^3}{3}\implies V=\cfrac{4\pi }{3}r^3
\\\\\\
\cfrac{dV}{dt}=\cfrac{4\pi }{3}\cdot \stackrel{chain~rule}{3r^2\cdot \cfrac{dr}{dt}}\quad 
\begin{cases}
\frac{dr}{dt}=50\\
r=6
\end{cases}\implies \cfrac{dV}{dt}=\cfrac{4\pi }{3}\cdot 3(6)^2\cdot 50\\\\\\ \cfrac{dV}{dt}=\stackrel{\frac{cm^3}{s}}{7200\pi}$

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

c)

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

Prove that the value of the expression does not depend on a: a−(6a−(5a−8))

djyliett [7]

Step-by-step explanation:

a−(6a−(5a−8))

a-(6a-5a+8)

a-6a+5a-8

6a-6a-8

0-8

=-8

so the answer doesn't depend on a

3 0

3 years ago

Esmeralda rent a car from a company that race cars by an hour she has a pair nishal fee of $52 and then they charged her $8 per

tatuchka [14]

Answer:

11 hours is the most she can afford

Step-by-step explanation:

7 0

3 years ago