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Delvig [45]
3 years ago
5

100 POINTS AND BRAINLIEST ANSWER

Mathematics
2 answers:
Karo-lina-s [1.5K]3 years ago
8 0

Answer:

Instantaneous rate of change = -13

Equation of tangent line : y= -2x +6

Step-by-step explanation:

the instantaneous rate of change of the function f(x) = 6x^2 - x at the point (-1,7)

To find instantaneous rate of change we take derivative

f(x) = 6x^2 - x

f'(x) = 12x-1

Now we plug in  -1 for x

f'(-1) = 12(-1)-1=-13

Instantaneous rate of change = -13

the equation of the line tangent to the function f(x) at the point (1,4)

f(x) = -x^2 + 5

To find slope of tangent line take derivative

f'(x) = -2x

Plug in 1 for x . given point is (1,4)

f'(1) = -2(1)=-2

so  slope = -2

use equation y-y1=m(x-x1)

y-4=-2(x-1)

y-4=-2x+2

Add 4 on both sides

y= -2x +6

polet [3.4K]3 years ago
5 0

Answer:

Just took the test,  

1)d, 3

2)a, -13

3) c, y=-2x+6

4) c, f'(a)=8a+2

Step-by-step explanation:

For #3, graph f(x)=-x^2+5. Next graph y=-2x+6. You will see that the added equation touches at exactly the point (1,4). this make it the tangent line equation

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<u>Pressure</u><u> </u><u>is</u><u> </u><u>8</u><u>0</u><u>,</u><u>0</u><u>0</u><u>0</u><u> </u><u>pascals</u><u>.</u>

Step-by-step explanation:

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Pressure:

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<h3><u>Answer:</u></h3>

\boxed{\boxed{\pink{\bf \leadsto Hence \ option\ [d]\ \bigg(y =  \dfrac{5}{2}x + 5\bigg) \ is \ correct  }}}

<h3><u>Step-by-step explanation:</u></h3>

Here from the given graph we can see that the graph the graph intersects x axis at (2,0) and y axis at (5,0). On seeing options it's clear that we have to use Slope intercept form . Which is :-

\large\boxed{\orange{\bf y = mx + c} }

We know that slope is \tan\theta. So here slope will be ,

\red{\implies slope = \tan\theta} \\\\\implies slope =\dfrac{PERPENDICULAR}{BASE} \\\\\bf \implies slope =\dfrac{5}{2}

Hence the slope is 5/2 . And here value of c will be 5 since it cuts y axis at (5,0).

\purple{\implies y = mx + c }\\\\\implies y = \dfrac{5}{2}x + 5 \\\\\underline{\boxed{\red{\tt \implies y = \dfrac{5}{2}x + 5 }}}

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Step-by-step explanation:

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3y''-6y'+6y=e*x sexcx
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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
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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

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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
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Therefore the particular solution is

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so that the general solution to the ODE is

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