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3 years ago

The slant asymptote of f(x)=

Mathematics

1 answer:

nekit [7.7K]3 years ago

7 0

Answer: It is true, I explained it down below in the picture.

Step-by-step explanation:

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Can some one-pls help me <br><br><br> -kaileyyy

Dmitriy789 [7]

It should be 4 qt + 1 qt is 1 gallon, I think

8 0

3 years ago

Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.

Ann [662]

Note that if ${x_2}'=x_1$ , then ${x_2}''={x_1}'$ , and so we can collapse the system of ODEs into a linear ODE:

${x_2}''=3{x_2}'-x_2+e^t$
${x_2}''-3{x_2}'+x_2=e^t$

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

$r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0$

so that the characteristic solution is

${x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}$

Now let's suppose the particular solution is ${x_2}_p=ae^t$ . Then

${x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t$

and so

$ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1$

Thus the general solution for $x_2$ is

$x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t$

and you can find the solution $x_1$ by simply differentiating $x_2$ .

7 0

3 years ago

In 1985 there were 275 cell phone subscribers in the small town of China grove. Subscribers increased by 40% each year after 198

emmasim [6.3K]

1985year ->275cell phones->100%
1994year -> x cell phones ->100%+360%
40% • (1994-1985)= 40% • 9= 360%

275 cell phone 100%
X cell phones. 100+360%=460%
x=275•460%/100%
x=12420/100
x=1265
99% sure this is a correct answer

x=275 • 460%/100%
x=12,420/100
x=1265

1994-1985= 9 yrs • 40%= 360%

5 0

3 years ago

An exit poll in an election is a survey taken of voters just after they have voted. One major use of exit polls has been so that

3241004551 [841]

Answer:

P(A) = 0.39

Step-by-step explanation:

We are given;

P(W|A) = 0.7

P(W|A^c ) = 0.3

We are told that 60% of the respondents said they voted for A. Thus;

P(A|W) = 60% = 0.6

Now, using the principle of drawing lots, we can be able to find the probability of the event that they are willing to participate in the exit poll which is P(W).

Thus;

P(W) = [P(W|A) × P(A)] +[P(W∣A^c) × P(A^c)]

Now, P(A^c) can be expressed as 1 - P(A)

Thus, we now have;

P(W) = [P(W|A) × P(A)] + [P(W∣A^c) × (1 - P(A)]

Plugging in the relevant values gives;

P(W) = 0.7P(A) + 0.3(1 - P(A))

P(W) = 0.7P(A) + 0.3 - 0.3P(A)

P(W) = 0.3 + 0.4P(A)

Now,using Baye's theorem, we can find an expression for P(A|W)

Thus;

P(A|W) = [P(A ∩ W)]/P(W)

This can be further expressed as;

P(A|W) = [P(A) × P(W|A)]/P(W)

Plugging in relevant values, we have;

0.6 = 0.7P(A)/(0.3 + 0.4P(A))

Cross multiply to get;

0.6(0.3 + 0.4P(A)) = 0.7P(A)

0.18 + 0.24P(A) = 0.7P(A)

0.18 = 0.7P(A) - 0.24P(A)

0.46P(A) = 0.18

P(A) = 0.18/0.46

P(A) = 0.39

7 0

3 years ago

Phil made 25% more cookies than Matt

pochemuha

72.00/0.25 = 288 then 288/2 = 144 and you would then do 1.25 x 144 = 180,
Phil baked 180 cookies

6 0

2 years ago