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2 years ago

1.7.3 Quiz: Intersecting Lines and Proofs

Mathematics

2 answers:

faltersainse [42]2 years ago

7 0

Answer:

Perpendicular lines.

Perpendicular lines are lines that intersect at a right (90 degrees) angle.

GalinKa [24]2 years ago

6 0

Perpendicular lines are 2 lines that intersect each other creating a 90 degree angle

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What's 1.8499 x 10^9 in standard notation

maxonik [38]

you just have to move the decimal point 9 places to the right thats an easy trick to remember 1849900000

5 0

3 years ago

PLEASE HELP ASAP!!!

barxatty [35]

Answers:

<u>24000 dollars</u> invested at 4%
<u>18000 dollars</u> was invested at 7%

======================================================

Work Shown:

x = amount invested at 4%

If she invests x dollars at 4%, then the rest (42000-x) must be invested at the other rate of 7%

She earns 0.04x dollars from that first account and 0.07(42000-x) dollars from the second account

This means we have

0.04x+0.07(42000-x)

0.04x+0.07*42000-0.07x

0.04x+2940-0.07x

-0.03x+2940

This represents the total amount of money earned after 1 year.

We're told the amount earned in interest is $2220, so we can say,

-0.03x+2940 = 2220

-0.03x = 2220-2940

-0.03x = -720

x = -720/(-0.03)

x = 24000 dollars is the amount invested at 4%

42000-x = 42000-24000 = 18000 dollars was invested at 7%

----------------------

As a check, we can see that

18000+24000 = 42000

and also

0.04x = 0.04*24000 = 960 earned from the first account

0.07*18000 = 1260 earned from the second account

1260+960 = 2220 is the total interest earned from both accounts combined

This confirms our answers.

7 0

3 years ago

In which equation does x = 15?

S_A_V [24]

To help you with it, plug-in 15 in each equation and see what you get!

8 0

2 years ago

Read 2 more answers

PLEASEE HELPP!!!

Mandarinka [93]

Answer:

AB/DE=BC/EF=AC/DF

Step-by-step explanation:

Corresponding segments are designated by letters in corresponding positions in the triangle names. For example, of one segment is designated using the 1st and 2nd letters of one triangle name (such as AB), then the corresponding segment is designated using the 1st and 2nd letters of the other triangle name (such as DE).

Corresponding segments are proportional in length. Corresponding angles are congruent.

6 0

3 years ago

Read 2 more answers

Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7

Digiron [165]

$y=x^r$
$\implies r(r-1)x^r+6rx^r+4x^r=0$
$\implies r^2+5r+4=(r+1)(r+4)=0$
$\implies r=-1,r=-4$

so the characteristic solution is

$y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}$

As a guess for the particular solution, let's back up a bit. The reason the choice of $y=x^r$ works for the characteristic solution is that, in the background, we're employing the substitution $t=\ln x$ , so that $y(x)$ is getting replaced with a new function $z(t)$ . Differentiating yields

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)$

Now the ODE in terms of $t$ is linear with constant coefficients, since the coefficients $x^2$ and $x$ will cancel, resulting in the ODE

$\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t$

Of coursesin, the characteristic equation will be $r^2+6r+4=0$ , which leads to solutions $C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}$ , as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If $z_1,z_2$ are the solutions to the characteristic equation of the ODE in terms of $z$ , then we can find another of the form $z_p=u_1z_1+u_2z_2$ where

$u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$
$u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$

where $W(z_1,z_2)$ is the Wronskian of the two characteristic solutions. We have

$u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t$

$u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t$

$\implies z_p=u_1z_1+u_2z_2$
$\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t$

and recalling that $t=\ln x\iff e^t=x$ , we have

$\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x$

4 0

2 years ago