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san4es73 [151]
3 years ago
8

Please help asap! thank you

Mathematics
2 answers:
nasty-shy [4]3 years ago
6 0
The HTTP 404, 404 Not Found, 404, 404 Error, Page Not Found, File Not Found, or Server Not Found
ololo11 [35]3 years ago
3 0

Answer:

w = ( 6x^5 - 18x^4 - 30x^3 + 28x^2 - 26x + 40 ) / ( x-4 )

Step-by-step explanation:

A = L x W

A = 6x^5 - 18x^4 - 30x^3 + 28x^2 - 26x + 40

L = ( x-4 )

W = A / L

W = ( 6x^5 - 18x^4 - 30x^3 + 28x^2 - 26x + 40 ) / ( x-4 )

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Please help i don’t understand
Natali [406]

Answer:

V= about 4,222.3 units

Step-by-step explanation:

d=24 so r=12 d/2=r

V= pi r sq times h/3

V=pi 12 sq times 28/3

V= pi 144 (28/3)

V= about 4,222.3

3 0
3 years ago
Using point-slope form, write an equation of the line that passes through (2,-1) and has a slope of -5
expeople1 [14]
Y – y1 = m(x – x1)

y-(-1)=-5(x-2)

Y+1=-5(x-2)
5 0
3 years ago
The length of a rectangular prism is 1.5 feet. Its width is 7 feet and its height is 6.2 feet. So the volume of the prism is 65.
Alexxandr [17]

7 * 6.2 * 1.5 = 65.1

the answer is True

5 0
3 years ago
Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).
I am Lyosha [343]
Take the homogeneous part and find the roots to the characteristic equation:

y''+4y=0\implies r^2+4=0\implies r=\pm2i

This means the characteristic solution is y_c=C_1\cos2x+C_2\sin2x.

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form y_p=ax\cos2x+bx\sin2x. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With y_1=\cos2x and y_2=\sin2x, you're looking for a particular solution of the form y_p=u_1y_1+u_2y_2. The functions u_i satisfy

u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx

where W(y_1,y_2) is the Wronskian determinant of the two characteristic solutions.

W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2

So you have

u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx
u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x

u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx
u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x

So you end up with a solution

u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x

but since \cos2x is already accounted for in the characteristic solution, the particular solution is then

y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x

so that the general solution is

y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x
7 0
3 years ago
Jodi is cutting out pieces of paper that measure 8 1/2 inches by 11 inches from a larger sheet of paper that has an area of 1000
jolli1 [7]
8 1/2 or 8.5 x11 = 93.5 square inches thus,
1000 square inches divided by 93.5 square inches
1000/93.5 ~10.695
that mean that Jodi can cut out about 10 pieces of paper measuring 8 1/2 x 11 inches.
4 0
3 years ago
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