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3 years ago

Equation of the tangent line

Mathematics

1 answer:

GalinKa [24]3 years ago

5 0

Find the eqn. of the tangent line to the curve of f(x) = x^2 + 5x -5 at (0,-5).

Differentiate f(x) to obtain an expression for the derivative (slope of the tangent line):

f '(x) = 2x + 5

Subst. 0 for x here: f '(0) = 2(0) + 5 = 5 (at the point (0, -5))

Use the point-slope equation of a str. line to find the eqn of the tan. line:

y-k = m(x-h), where (h,k) is a point on the line and m is the slope:

y - [-5] = 5(x-0), or y+5 = 5x. Then y = 5x - 5 is the eqn. of the TL to the given curve at (0,-5).

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2) Prove that these two lines are parallel:

borishaifa [10]

Answer:

if the two lines have the same slope that means they are parallel

3 0

3 years ago

Please help me with precalculus??<br>linear and angular speed

Anit [1.1K]

13)

there are 2π radians in 1 revolution, and there are 60 seconds in 1 minute, so keeping that in mind, then,

$\bf \cfrac{4\underline{\pi} }{5~\underline{s}}\cdot \cfrac{rev}{2\underline{\pi} }\cdot \cfrac{60~\underline{s}}{min}\implies \cfrac{4\cdot 60~rev}{5\cdot 2~min}\implies \cfrac{240~rev}{10~min}\implies 24\frac{rev}{min}$

14)

$\bf \textit{linear velocity}\\\\
v=rw\quad 
\begin{cases}
r=radius\\
w=angular~speed\\
----------\\
v=32\frac{m}{sec}\\
w=100\frac{rev}{min}
\end{cases}\\\\
-------------------------------\\\\
\textit{let's convert \underline{w} to }\frac{radians}{sec}$

$\bf \cfrac{100~\underline{rev}}{\underline{min}}\cdot \cfrac{2\pi }{\underline{rev}}\cdot \cfrac{\underline{min}}{60~sec}\implies \cfrac{100\cdot 2\pi }{60~sec}\implies \cfrac{10\pi }{3~sec}\implies \cfrac{10\pi }{3}\frac{radians}{sec}\\\\
-------------------------------\\\\
v=rw\implies \cfrac{v}{w}=r\implies \cfrac{\frac{30~m}{sec}}{\frac{10\pi }{3~sec}}\implies r=\cfrac{30~m}{\underline{sec}}\cdot \cfrac{3~\underline{sec}}{10\pi }
\\\\\\
r=\cfrac{90}{10\pi }m$

15)

what is the radians per seconds "w" in revolutions per minute? just another conversion like in 13)

$\bf \cfrac{\underline{\pi} }{3~\underline{sec}}\cdot \cfrac{rev}{2\underline{\pi }}\cdot \cfrac{60~\underline{sec}}{min}\implies \cfrac{60 ~rev}{3\cdot 2 ~min}\implies \cfrac{60 ~rev}{6 ~min}\implies 10\frac{rev}{min}$

4 0

3 years ago

Can u help this helpless 9th grader ?

nexus9112 [7]

Answer:

y = 4x + 32

Step-by-step explanation:

Given: Temperature at sunrise = 32°F

y is the temperature after x hours of sunrise

Temperature change is 4°F/hour.

The equation of a line in slope intercept form is given by

y = mx +c

where m is the slope, which is the change in value of y to the value of x

and c is the value of y at x = 0

At sunrise, x =0 and y =32

∴ c = 32

Also we see that the change in temperature to that of time is 4/1 = 4

∴ m = 4

Substituting the values of m and c in the equation of line, we have

y = 4x +32

7 0

3 years ago

Read 2 more answers

The accompanying data on x = current density (mA/cm2) and y = rate of deposition (m/min)μ appeared in a recent study.

gtnhenbr [62]

Answer:

a) $r=\frac{4(333)-(200)(5.37)}{\sqrt{[4(12000) -(200)^2][4(9.3501) -(5.37)^2]}}=0.9857$

The correlation coefficient for this case is very near to 1 so then we can ensure that we have linear correlation between the two variables

b) $m=\frac{64.5}{2000}=0.03225$

Now we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{200}{4}=50$

$\bar y= \frac{\sum y_i}{n}=\frac{5.37}{4}=1.3425$

$b=\bar y -m \bar x=1.3425-(0.03225*50)=-0.27$

So the line would be given by:

$y=0.3225 x -0.27$

Step-by-step explanation:

Part a

The correlation coeffcient is given by this formula:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

For our case we have this:

n=4 $\sum x = 200, \sum y = 5.37, \sum xy = 333, \sum x^2 =12000, \sum y^2 =9.3501$

$r=\frac{4(333)-(200)(5.37)}{\sqrt{[4(12000) -(200)^2][4(9.3501) -(5.37)^2]}}=0.9857$

The correlation coefficient for this case is very near to 1 so then we can ensure that we have linear correlation between the two variables

Part b

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=12000-\frac{200^2}{4}=2000$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=333-\frac{200*5.37}{4}=64.5$

And the slope would be:

$m=\frac{64.5}{2000}=0.03225$

Now we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{200}{4}=50$

$\bar y= \frac{\sum y_i}{n}=\frac{5.37}{4}=1.3425$

And we can find the intercept using this:

$b=\bar y -m \bar x=1.3425-(0.03225*50)=-0.27$

So the line would be given by:

$y=0.3225 x -0.27$

4 0

3 years ago

i really need help with my math assignment it’s normal distributions and i need this to graduate tomorrow if anyone can help it

Vika [28.1K]

Where’s the work you need help with :)

7 0

3 years ago