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lapo4ka [179]
3 years ago
15

How do i help people by answering there question

Mathematics
1 answer:
Fudgin [204]3 years ago
3 0
What do you mean by this?
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How to solve this math -9(-6w+3u-4)
Goryan [66]
-54w-27u+36 because -9 times 6 = 54 and -9 times 3 = 27 and -9 times -4 = 36 so -54w -27 u + 36
3 0
3 years ago
This question worth 38 points to whoever answers it :)
Paraphin [41]
Volumeprism=volumecube times number of cubes
vprism/vcube=number of cubes

vprism=12*5*4.25=255
vcube=side^3=(1/4)^3=1/64

vprism/vcube=255/(1/64)=255*64=16320 cubes

4 0
3 years ago
Read 2 more answers
Consider 7×10^3. Write a pattern to find the value of the expression
kkurt [141]

Answer:

7*10*10*10 = 7000

7*10 = 70

70*10 = 700

700*10 = 7000

Step-by-step explanation:

The given expression is:

7*10^3

Here 10^3 means that 10 will be multiplied 3 times:

7*10*10*10 = 7000

How did we get 7000?

7*10*10*10 = 7000

7*10 = 70

70*10 = 700

700*10 = 7000

We can also say that there are 7  1000s in 7000....

5 0
3 years ago
Read 2 more answers
3y''-6y'+6y=e*x sexcx
Simora [160]
From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

3r^2-6r+6=0\iff r^2-2r+2=0

which has roots at r=1\pm i. This admits the two fundamental solutions

y_1=e^x\cos x
y_2=e^x\sin x

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the fundamental solutions. We have

W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}

and so

u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx
u_1=\dfrac13\ln|\cos x|

u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx
u_2=\dfrac13x

Therefore the particular solution is

y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x

so that the general solution to the ODE is

y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x
7 0
3 years ago
Place the steps for finding f(x) in the correct order.
Naily [24]
0
The answer is C
Ik this bc
Someone told me
8 0
2 years ago
Read 2 more answers
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