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3 years ago

14

Carlos drives 2 times the hours that patricia does. carlos and patricia drive for a total of 9 hours. how long did each person d

rive?

1 answer:

suter [353]3 years ago

4 0

Patricia drove for 3 hours. Carlos drove for twice that, 6 hours. 6+3=9

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Use the point (-6,-8) to give the six trigonometric functions

vfiekz [6]

You can't exactly get the trigonometric functions using a point unless it's a vector,

so tanx = -8/-6 = 4/3 and from it we can get the other functions

sinx = -4/5
cosx = -3/5
secx = -5/3
cscx = -5/4
cotx = 3/4

3 0

3 years ago

Please do answer: <br><br>Simplify the expression---5x-2y + 3x + 2y<br><br>(100 points)

netineya [11]

5x-2y+3x+2y

Combine like terms

5x+3x-2y+2y
(5+3)x+(-2+2)y
8x+0
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2 years ago

Read 2 more answers

2y=8x What is the slope intercept form, slope, and y-intercept

kaheart [24]

Y-intercept: 0
Slope: 4

Explanation
Y-intercept is the number without the x which in this case is no number so it’s 0 so that’s where the point in the y axis goes

Slope is rise over run so if the equation is 2y = 8x then we can simplify this by dividing both sides by 2 which would be y = 4x so slope is 4/x. Which is basically 4/1 because the x means over one so up 4 units and to the right once

Hope this helps

3 0

2 years ago

A tailor earns (12y² + y – 35) pesos for working (3y – 5) hours. How much does he earn per hour?

Natalka [10]

Answer:

4y+7

Step-by-step explanation:

(12y² + y – 35)/(3y – 5)

=(3y-5)(4y+7)/(3y-5) (factorization)

=4y + 7

Hope this helped!

4 0

3 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