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4 years ago

12

When constructing inscribed polygons and perpendicular lines how are the steps similar

1 answer:

bagirrra123 [75]4 years ago

8 0

When constructing inscribed polygons and perpendicular lines, how are the steps similar?
<span>There are no similarities.</span>

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If the terminal point of ø is (1,0), what is tan ø?

charle [14.2K]

Given:

The terminal point of $\phi$ is (1,0).

To find:

The value of $\tan\phi$ .

Solution:

If the terminal point of $\theta$ is (x,y), then

$\tan \theta=\dfrac{y}{x}$

It is given that the terminal point of $\phi$ is (1,0).

Here, x-coordinate is 1 and the y-coordinate is 0. Using the above formula, we get

$\tan \phi=\dfrac{0}{1}$

$\tan \phi=0$

Therefore, the value of $\tan \phi$ is 0.

4 0

3 years ago

-1x= -8-2x<br><br> what is x <br> pls i need help

UkoKoshka [18]

Answer:

x = -8

General Formulas and Concepts:

Order of Operations: BPEMDAS

Step-by-step explanation:

<u>Step 1: Define</u>

-1x = -8 - 2x

<u>Step 2: Solve for </u><em><u>x</u></em>

Simplify: -x = -8 - 2x
Add 2x on both sides: x = -8

<u>Step 3: Check</u>

<em>Plug in x to verify it's a solution.</em>

Substitute: -1(-8) = -8 - 2(-8)
Multiply: 8 = -8 + 16
Add: 8 = 8

3 0

3 years ago

Read 2 more answers

Which solution is graphed on the number line?

zvonat [6]

Answer:

x < 1

Step-by-step explanation:

6 0

3 years ago

Find a second solution y2(x) of<br> x^2y"-3xy'+5y=0; y1=x^2cos(lnx)

rosijanka [135]

We can try reduction order and look for a solution $y_2(x)=y_1(x)v(x)$ . Then

$y_2=y_1v\implies{y_2}'=y_1v'+{y_1}'v\implies{y_2}''=y_1v''+2{y_1}'v+{y_1}''v$

Substituting these into the ODE gives

$x^2(y_1v''+2{y_1}'v+{y_1}''v)-3x(y_1v'+{y_1}'v)+5y_1v=0$

$x^2y_1v''+(2x^2{y_1}'-3xy_1)v'+(x^2{y_1}''-3x{y_1}'+5y_1)v=0$

$x^4\cos(\ln x)v''+x^3(\cos(\ln x)-2\sin(\ln x))v'=0$

which leaves us with an ODE linear in $w(x)=v'(x)$ :

$x^4\cos(\ln x)w'+x^3(\cos(\ln x)-2\sin(\ln x))w=0$

This ODE is separable; divide both sides by the coefficient of $w'(x)$ and separate the variables to get

$w'+\dfrac{\cos(\ln x)-2\sin(\ln x)}{x\cos(\ln x)}w=0$

$\dfrac{w'}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}$

$\dfrac{\mathrm dw}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}\,\mathrm dx$

Integrate both sides; on the right, substitute $u=\ln x$ so that $\mathrm du=\dfrac{\mathrm dx}x$ .

$\ln|w|=\displaystyle\int\frac{2\sin u-\cos u}{\cos u}\,\mathrm du=\int(2\tan u-1)\,\mathrm du$

Now solve for $w(u)$ ,

$\ln|w|=-2\ln(\cos u)-u+C$

$w=e^{-2\ln(\cos u)-u+C}$

$w=Ce^{-u}\sec^2u$

then for $w(x)$ ,

$w=Ce^{-\ln x}\sec^2(-\ln x)$

$w=C\dfrac{\sec^2(\ln x)}x$

Solve for $v(x)$ by integrating both sides.

$v=\displaystyle C_1\int\frac{\sec^2(\ln x)}x\,\mathrm dx$

Substitute $u=\ln x$ again and solve for $v(u)$ :

$v=\displaystyle C_1\int\sec^2u\,\mathrm du$

$v=C_1\tan u+C_2$

then for $v(x)$ ,

$v=C_1\tan(\ln x)+C_2$

So the second solution would be

$y_2=x^2\cos(\ln x)(C_1\tan(\ln x)+C_2)$

$y_2=C_1x^2\sin(\ln x)+C_2x^2\cos(\ln x)$

$y_1(x)$ already accounts for the second term of the solution above, so we end up with

$\boxed{y_2=x^2\sin(\ln x)}$

as the second independent solution.

6 0

4 years ago

(Middle school math) please help, thank you!

alexira [117]

Answer:

Length of the fence will be = 282.84 feet

Step-by-step explanation:

The Blackburn family has a square field to keep their cattle. Area of the field is $40000 square feet.

Then one side of the square field will be = √Area = √40000 = 200 feet

Since diagonal of a square is = √2(side)² = side√2 = 200√2 = 200×1.41 = 282.84 feet.

Therefore length of the fence will be = 282.84 ft.

3 0

3 years ago

Read 2 more answers