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3 years ago

14

Two parallel lines are crossed by a traversal. What is the value of h?

1 answer:

maxonik [38]3 years ago

8 0

Answer:

h=60

Step-by-step explanation:

The angle that is 120 degrees and the angle that is h degrees are called same side interior angles

Same side interior angles are supplementary, meaning that they add to 180 degrees

To find h we can use Measures addition postulate to find the unknown value h

h+120=180

h=60

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Write a expression the product of eght and seven less than a number

MrRissso [65]

Answer:

Step-by-step explanation:

The question says that you are multiplying 8 and something together. So to start with, it looks like this.

8*something.

Now you have to get 7 less a number which is 7 - x

So something is 7 - x

8(7 - x) is your answer.

7 0

3 years ago

74$ for working 6 hours.what is the hourly pay rate in hours per dollar

fgiga [73]

Answer:

.081 hours per dollar

$12.333 dollars per hour

Step-by-step explanation:

I'm a bit confused whether you just want hours per dollar or also dollars per hour, so I'll answer both.

First, for the dollars per hour, you want to do 74 divided by 6 equals 12 1/3. So each hour you get paid $12.333

Fort the hours per dollar, you do 6 divided by 74 equals .081.

5 0

3 years ago

John works two jobs. He earns $10 an hour babysitting his neighbor's children and $15 an hour mowing lawns in his neighborhood.

Jlenok [28]

Add up all the numbers

7 0

2 years ago

Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.

jarptica [38.1K]

$y''-y'+y=\sin x$

The corresponding homogeneous ODE has characteristic equation $r^2-r+1=0$ with roots at $r=\dfrac{1\pm\sqrt3}2$ , thus admitting the characteristic solution

$y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x$

For the particular solution, assume one of the form

$y_p=a\sin x+b\cos x$

${y_p}'=a\cos x-b\sin x$

${y_p}''=-a\sin x-b\cos x$

Substituting into the ODE gives

$(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x$

$-b\cos x+a\sin x=\sin x$

$\implies a=1,b=0$

Then the general solution to this ODE is

$\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}$

$y''-3y'+2y=e^x\sin x$

$\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2$

$\implies y_c=C_1e^x+C_2e^{2x}$

Assume a solution of the form

$y_p=e^x(a\sin x+b\cos x)$

${y_p}'=e^x((a+b)\cos x+(a-b)\sin x)$

${y_p}''=2e^x(a\cos x-b\sin x)$

Substituting into the ODE gives

$2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x$

$-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x$

$\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12$

so the solution is

$\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}$

$y''+y=x\cos(2x)$

$r^2+1=0\implies r=\pm i$

$\implies y_c=C_1\cos x+C_2\sin x$

Assume a solution of the form

$y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)$

${y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)$

Substituting into the ODE gives

$(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)$

$-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)$

$\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49$

so the solution is

$\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}$

7 0

3 years ago

What is 1 and 3/10 plus 3 and 1/3

Mamont248 [21]

Answer:

4 19/30

Step-by-step explanation:

1 3/10 + 3 1/3 = 139/30 or 4.63333...

Simplyfiy 139/30 to 4 19/30

3 0

3 years ago