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3 years ago

Anybody know this!? Help please ..

Mathematics

1 answer:

lubasha [3.4K]3 years ago

4 0

The slopes of perpendicular lines are negative reciprocal to each other
ex. the negative reciprocal to 5/2 is -2/5

once you find the perpendicular slope, you plug in the point and slope into the point-slope form of a line and then rearrange it into the slope-intercept form to get your answer

point slope
$y-y_1 = m(x-x_1)$

slope intercept form
$y=mx+b$

You might be interested in

Amirah is 4 years older than Priya and Shirley is

jenyasd209 [6]

Answer:

Priyah = 15

Amirah = 19

Shirley = 13

Step-by-step explanation:

Let A represents Amirah, P Priyah and S, Shirley.

So, we have:

$A = 4 + P$ --- 4 years older

$S = P - 2$ --- 2 years younger

$A + P + S = 47$

Required

Find their ages

Substitute 4 + P for A and P - 2 for S in $A + P + S = 47$

$4 + P + P + P - 2 = 47$

$4 + 3P - 2 = 47$

Collect Like Terms

$3P = 47 - 4+ 2$

$3P = 45$

Divide both sides by 3

$P = \frac{45}{3}$

$P = 15$

Substitute 15 for P in $A = 4 + P$ and $S = P - 2$

$A = 4 + 15$

$A = 19$

$S = 15 - 2$

$S = 13$

8 0

3 years ago

Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

The polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

$\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)$

with

$\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}$

for some c in the interval (-x, x)

In the particular case f

f(x)=cos(x)

we have

$\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)$

therefore

$\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0$

and the polynomial approximation of T5(x) of cos(x) would be

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

$\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}$

for some c in (-x,x). So

$\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}$

and we must find the values of x for which

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652$

Working this inequality out, we find

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$