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4 years ago

Find the missing lengths: HI=7 and LH=9, find LK.

Mathematics

1 answer:

Katarina [22]4 years ago

8 0

Answer:

LK=12

Step-by-step explanation:

From the given figure, HI=7 and LH=9, and we know that

$(KH)^{2}=(IH)(HL)$

$(KH)^{2}=7{\times}9$

$(KH)^{2}=63$ (1)

Now, from ΔKHL, we have

$(OH)^{2}=(OK)(OL)$

and from ΔOKH,

$(KH)^{2}=(OH)^{2}+(OK)^2$

$(OK)^2=(KH)^2-(OH)^2$

$(OK)^2=63-(OK)(OL)$

$(OK)^2+(OK)(OL)=63$

$OK(OK+OL)=63$

$OK(KL)=63$ (2)

Also, ΔIKL is similar to ΔOHL, therefore

$\frac{OL}{KL}=\frac{9}{16}$

$\frac{KL-OK}{KL}=\frac{9}{16}$

$1-\frac{OK}{KL}=\frac{9}{16}$

$\frac{7}{16}=\frac{OK}{KL}$

Using equation (2), we get

$\frac{7}{16}=\frac{63}{(KL)^2}$

$(KL)^2=\frac{1008}{7}$

$(KL)^2=144$

$KL=12$

Thus, the value of LK is 12.

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A 20-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 30 o

Dafna1 [17]

Answer:

$i(t)=(2/3)(1-e^{-300t})$

Step-by-step explanation:

Before we even begin it would be very helpful to draw out a simple layout of the circuit. Then we go ahead and apply kirchoffs second law(sum of voltages around a loop must be zero) on the circuit and we obtain the following differential equation,

$-V +Ldi/dt+Ri=0$

where V is the electromotive force applied to the LR series circuit, Ldi/dt is the voltage drop across the inductor and Ri is the voltage drop across the resistor. we can re write the equation as,

$di/dt+Ri/L=V/L$

Then we first solve for the homogeneous part given by,

$di/dt+Ri/L=0$

we obtain,

$i(t)_{h} =I_{max}e^{-Rt/L}$

This is only the solution to the homogeneous part, The final solution would be given by,

$i(t)=i(t)_{h} +c$

where c is some constant, we added this because the right side of the primary differential equation has a constant term given by V/R. We put this in the main differential equation and obtain the value of c as c=V/R by comparing the constants on both sides.if we put in our initial condition of i(0)=0, we obtain $I_{max} =V/R$ , so the overall equation becomes,

$I(t)=(V/R)(1-e^{-Rt/L})$

where if we just plug in the values given in the question we obtain the answer given below,

$i(t)=(2/3)(1-e^{-300t})$

5 0

4 years ago

Find all the solutions of the given equations in the interval [0,2pi) tan^3x=tanx and cos 3x=-cos3x

slamgirl [31]

$\tan^3x=\tan x$
$\tan^3x-\tan x=0$
$\tan x(\tan^2x-1)=0$
$\tan x(\tan x-1)(\tan x+1)=0$
$\begin{cases}\tan x=0\\\tan x-1=0\\\tan x+1=0\end{cases}$

$\tan x=\dfrac{\sin x}{\cos x}=0\implies \sin x=0\implies x=0,\pi$

$\tan x-1=0\iff\tan x=1\implies x=\dfrac\pi4,\dfrac{5\pi}4$

$\tan x+1=0\iff\tan x=-1\implies x=\dfrac{3\pi}4,\dfrac{7\pi}4$

- - -

$\cos3x=-\cos3x$
$2\cos3x=0$
$\cos3x=0\implies 3x=\dfrac\pi2,3x=\dfrac{3\pi}2\implies x=\dfrac\pi6,x=\dfrac\pi2$

This doesn't account for all the solutions, however; there are some values of $x$ that push $3x$ outside the interval $[0,2\pi)$ , so let's take a few more:

$3x=\dfrac{5\pi}2\implies x=\dfrac{5\pi}6$
$3x=\dfrac{7\pi}2\implies x=\dfrac{7\pi}6$
$3x=\dfrac{9\pi}2\implies x=\dfrac{9\pi}6$
$3x=\dfrac{11\pi}2\implies x=\dfrac{11\pi}6$

We can stop there, since the next candidate gives

$3x=\dfrac{13\pi}2\implies x=\dfrac{13\pi}6>2\pi$

7 0

3 years ago

What is the amplitude of the graph below

Kay [80]

The amplitude is 3

for a sine graph, the amplitude is the distance up or down from the midway point. The equation to find the amplitude is (max-min) / 2.

(3 - (-3)) / 2 = 3

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3 years ago

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Shkiper50 [21]

X² - 5x - 14 = (x - 7)(x + 2)

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ZanzabumX [31]

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3 years ago

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