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2 years ago

6

Which similarity statement best fits this picture

1 answer:

saveliy_v [14]2 years ago

4 0

Second answer I think

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a line with the slope of 1/3 passes through the points of (6,0). what is the equation in slope intercept form?

sleet_krkn [62]

Answer:

y = 1/3x -2

Step-by-step explanation:

the equation for slope intercept form is y= mx + b.

m is the slope and b is where the line crosses the y-axis which is called the y-intercept.

to find the y-intercept plug in a coordinate on the line (6,0)

0 = 1/3(6) + b

then solve for b by first multiplying 1/3 by 6

0 = 2 + b

subtract 2 from both sides

-2 = b

and then you can plug b into the equation

y = 1/3x - 2

8 0

3 years ago

A movie at Walmart costs $14.97. Tax is 6%. What will the total be, including tax?

Ghella [55]

Answer:

multiply 14.97 to .6 and that's your answer

Step-by-step explanation:

3 0

3 years ago

One number is 5 more than the other. Their sum is 33. Find the numbers.

dimaraw [331]

Answer:

19,14

Step-by-step explanation:

x=5+y

x+ y= 33

we will solve using substitution

(5+y) +y= 33

5+y+y=33

5+2y=33

subtract 5 from both sides

2y=28

y=14

We got y now we will find what x is

for this, we will substitute y's value (14) into the first equation

x=5+(14)

x=19

we got x too

3 0

3 years ago

Someone deleted my questions- okay?<br><br> answer this then- 2/3-6/7+7/8

Afina-wow [57]

Answer:

8/5 + 6 2/7

Step-by-step explanation:

Step 1: Make sure the bottom numbers (the denominators) are the same. Step 2: Add the top numbers (the numerators), put that answer over the denominator. Step 3: Simplify the fraction (if needed) hope I 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