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frutty [35]
3 years ago
7

I can’t seem to find the value of n in this peoblem

Mathematics
2 answers:
umka2103 [35]3 years ago
5 0

Answer:

i think the value of n is 12

Annette [7]3 years ago
3 0

Answer:

9

Step-by-step explanation:

you can always use guess-and-check:

124 seems rather high

(6)P(2) = 6!/4! = 6x5 ≠ 72

(9)P(2) = 9!/7! = 9x8 = 72

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Alexeev081 [22]
Y-intercept (7,0) x-intercept (0,14) = f(x)= -2x+14
7 0
3 years ago
How many solutions does the equation have 4x + 3=2(2x +9)
lbvjy [14]
I guess no solution
4x +3 = 4X + 18
4X - 4X = 18 - 3
0 = 15
6 0
3 years ago
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
A community swimming pool is a rectangular prism that is 30 feet long, 12 feet wide, and 5 feet deep. The wading pool is half as
topjm [15]

Answer:

The volume of the community swimming pool is 4 times greaters than the volume of the wading pool.

Step-by-step explanation:

By definition of rectangular prism, we get the respective formulas for the volumes of the community swimming pool and the wadling pool, respectively:

Community swimming pool

V_{c} = l\cdot w\cdot h (1)

Wading pool

V_{w} = \left(\frac{1}{2}\cdot l \right)\cdot \left(\frac{1}{2}\cdot h\right)\cdot w (2)

Where:

l – Length of the swimming pool, measured in feet.

h – Depth of the swimming pool, measured in feet.

w – Width of the swimming pool, measured in feet.

V_{c} – Volume of the community swimming pool, measured in cubic feet.

V_{w} – Volume of the wading swimming pool, measured in cubic feet.

The ratio of the volume of the community swimming pool to the volume of the wadling pool is:

\frac{V_{c}}{V_{w}} = \frac{l\cdot w \cdot h}{\left(\frac{1}{2}\cdot l \right)\cdot \left(\frac{1}{2}\cdot h\right)\cdot w} (3)

\frac{V_{c}}{V_{w}} = \frac{1}{\frac{1}{4} }

\frac{V_{c}}{V_{w}} = 4

The volume of the community swimming pool is 4 times greaters than the volume of the wading pool.

8 0
2 years ago
Read 2 more answers
What shape can be formed by this net?
ivolga24 [154]
It’s triangular prism
5 0
3 years ago
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