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zvonat [6]
3 years ago
13

Find the derivative of the function by using the product rule. Simplify your answer.

Mathematics
1 answer:
OlgaM077 [116]3 years ago
5 0

Where is the question I dont see it sorry

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Simplify the expression by combining like terms. 7w+9+6X
Mars2501 [29]

Answer:

Step-by-step explanation:

the answer stay the same due to the fact there are no like terms

7w+9+6x

8 0
2 years ago
Read 2 more answers
If ABC and EDC are similar, what is the value<br> of x?<br> A
abruzzese [7]

Answer:

x = 15

Step-by-step explanation:

Since the triangles are similar then the ratios of corresponding sides are equal, that is

BC/DC = EC/AC , substitute values

3x×5/32 - 5x-5/56 ( cross- multiply )

56(3x - 5) = 32(5x - 5) ← distribute and simplify both sides

168x - 280 = 160x - 160 ( subtract 160x from both sides )

8x - 280 = - 160 ( add 280 to both sides )

8x = 120 ( divide both sides by 8 )

x = 15

3 0
3 years ago
Suppose A and B are independent events if P(A) = 0.4 And P(B) = 0.1, what is P(A'uB)? APEX
Aloiza [94]
For Independent Events, P(A) × P(B) = P(A∩B)

so we have, P(A∩B) = 0.4×0.1 = 0.04

P(A') = 1 - 0.4 = 0.6

This information can be represented on a Venn diagram as shown below

P(A'∪B) means the union of everything that is not A with everything that is B

P(A'∪B) = 0.06 + 0.54 + 0.04 = 0.64


5 0
2 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
2 years ago
What is 2 times 2? to the nerist nuber off the reel steel anser.
Llana [10]

Answer:

day also you have the same knowledg

6 0
2 years ago
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