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3 years ago

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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Find the pressure exerted by a force of 240 N on an area of 30 cm2. Find the answer in SI unit

NeTakaya

Answer:

Pressure is 80,000 pascals.

Step-by-step explanation:

Area:

$= { \bf{ 30 \times {10}^{ - 4} \: {m}^{2} }}$

Pressure:

${ \bf{pressure = \frac{force}{area} }} \\ \\ { \tt{ = \frac{240}{30 \times {10}^{ - 4} } }} \\ = { \tt{80000 \: pascals}}$

4 0

3 years ago

Find the equation of the graphed line.

belka [17]

<h3>Answer:</h3>

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

<h3>Step-by-step explanation:</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).

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

<h3>Hence option [ d ] is correct . </h3>

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2 years ago

Use the fact that the bacteria is doubling every five

victus00 [196]

Answer:

I think it's 1/16

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

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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bezimeni [28]

Answer:-46

Step-by-step explanation: just took test

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2 years ago