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Y_Kistochka [10]

3 years ago

12

The rectangle below represents the base of a rectangular prism. If the

1 answer:

Andrews [41]3 years ago

4 0

Answer:

64

Step-by-step explanation:

hiiii

to find volume the formula is

base * width * height.

so all we would have to do here is:

5 1/3 * 2 * 6=64.

hope this helped you :)

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Mike runs cross country. He runs 7 miles in 56

kondor19780726 [428]

It will take him 120 mins as he runs 8 minute miles

4 0

3 years ago

What is 2.99 as a fraction

Serga [27]

2.99

2 is a whole number

0.99 = 99/100

your fraction is:

mixed number = 2 99/100
improper fraction = 299/100 (multiply whole number to denominator & then add numerator)

hope this helps

8 0

3 years ago

Read 2 more answers

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 type of triangle is shown?

MAXImum [283]

Answer:

right triangle

Step-by-step explanation:

We know that the total angles of the triangle are 180 degrees:

(x) + (4x-10) + (2x+15) = 180

7x - 10 + 15 =180

7x + 5 = 180

7x = 180 - 5

7x = 175

x = 175÷ 5 = 25

Now we calculate each angle

x=25

4x-10= (4*25) -10=100-10=90

2x+15=(2*25)+15=50+15=65

Because we have a right angle, then the triangle is right triangle

6 0

2 years ago

Read 2 more answers

Please help with this 8x2+4y÷z when x=12, y=34, and z=2 I don't know this

natita [175]

8x12=96 4x34=136 136+96=232 232/2=116 answer=116

4 0

3 years ago