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Ira Lisetskai [31]
3 years ago
8

51) Eloise had to solve a system of inequalities that contained an exponential growth function and a linear function with positi

ve slope. She KNOWS that the two equations intersect at least once, and she was asked to state the interval where the linear > exponential. Which could be an answer to her system of inequalities? A) (-1, 3) B) (-1, [infinity]) C) (-[infinity], -1) D) (-[infinity], -1) ∪ (3, [infinity])
Mathematics
1 answer:
erik [133]3 years ago
8 0

Answer:

Answer is A. (-1, 3)

Refer below.

Step-by-step explanation:

If the line and exponential function intersect only once then the line is tangent to the exponential at that point and therefore is always is below the exponential except at the point of intersection where there are equal.

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Answer:

no bc the graph shows a slope of 12

Step-by-step explanation:

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Answer:

the estimated area of the fountain is 147.58 square yards and the estimated area of the path is 28.26 square yards

Step-by-step explanation:

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see i don't mind giving my points away..Can anyone help me solve this though for a brainlest and extra points?​
victus00 [196]

Answer:

a=17\sqrt{2}

Step-by-step explanation:

As we have 2 of the angles, we can find the third. The measure of the third angle is 45 degrees.

As this is the same measure as the other, that means that the side length will be the same.

Now was have two side lengths of 17 in a right triangle, so we can use the Pythagorean theorem to find a.

Recall that the pythagorean theorem states: a^2+b^2=c^2

In this case, a is 17, b is 17, and c is a

Knowing this, we can input our value into this formula and solve for a.

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3 years ago
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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
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Answer:

Step-by-step explanation:

34 to the 4

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