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3 years ago

Determine the distance between the line y = −4x − 7 and the point (1, 1). (Express the

1 answer:

5 0

$distance = | \frac{(- 4 \times 1) + (- 1 \times 1)-7}{ \sqrt{ {( - 4)}^{2} + {( - 1)}^{2} } } | \\= \frac{ - 12}{ \sqrt{17 } } \\ = \frac{ - 12 \sqrt{17} }{17} \\ = - 2.91$

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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

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

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

[-1.113826815, 1.113826815]

8 0

3 years ago

Find the measures of the missing angles Help pleaseee

kondaur [170]

Answer:

Nwm miałem to dawno powodzenia

6 0

3 years ago

Read 2 more answers

HELP PLEASEEE!!!! I START SCHOOL TOMORROW

ohaa [14]

Answer:

4. The range is -2 ≤ y ≤ 4

5. The domain is {-2 , -1 , 0 , 3.5 , 4.2}

6. The relation is not a function

Step-by-step explanation:

* Lets explain how to solve the problem

# Question 4:

- The range of the function is the values of all y (output) which is

corresponding to the values of all x (input)

- The figure is a stare, the values of x is from -4 to 4

∴ The domain is -4 ≤ x ≤ 4

- The corresponding values of y are from -2 to 4

∴The range is -2 ≤ y ≤ 4

# Question 5:

- The table represent a relation between x and y

- x is the input of the relation

- y is the output of the relation

∵ The input of the relation is called the domain of it

∴ The domain of the relation is all values of x in the table

- the values of x are -2 , -1 , 0 , 3.5 , 4.2

∴ The domain is {-2 , -1 , 0 , 3.5 , 4.2}

# Question 6:

- The relation can be a function if each ordered pair has a

unique value of x (input)

- That means every x (input) has only one y (output)

- Ex : The relation {(a , b) , (c , d) , (a , f)} is not a function because

the input a has two output b and f

∵ The relation is {(-4 , 0) , (-3 , 0) , (-2 , 1) , (1 , -2) , (-3 , 4)}

∵ The x = -3 has two values of y 0 and 4

∴ The relation is not a function

7 0

3 years ago

PLEASE HELP ASAP 50 POINTS! What correctly identifies the property shown?

mash [69]

Answer:

Commutative property

Step-by-step explanation:

Hope that helps!

5 0

3 years ago

Given f(x) = 10 - 16/x, find all c in the interval [2,8] that satisfy the mean value theorem

Katen [24]

F(x)=10-16/x
f'(x)=16/x^2
f'(c)=16/c^2
f(8)=10-16/8=10-2=8
f(2)=10-16/2=10-8=2
$f'(c)= \frac{f(b)-f(a)}{b-a} in (2,8),
 \frac{16}{c^2} = \frac{f(8)-f(2)}{8-2} ,
 \frac{16}{c^2} = \frac{8-2}{8-2}
$
4∈[2,8] is the required point.

8 0

4 years ago

Read 2 more answers