How I can answer this question, NO LINKS, if you answer correctly I will give u brainliest!

dimulka [17.4K]

A function is a relation in which the members of the domain (x-values) do not repeat. So, for every x-value there is only one y-value that corresponds to it.

4 0

2 years ago

PLEASE HELPPPPP

umka21 [38]

The answer is B I believe

4 0

2 years ago

Yolanda made $192 for 12 hours of work. At the same rate, how many hours would she have to work to make $272?

denpristay [2]

Answer:

17 hours

take $192 and divide it by 12 so you know how much they make per hour

$192 / 12 = 16

then you're going to take $272 and divide it by what you know now - how much they make per hour (this will get you how many hours they worked)

$272 / 16 = 17

6 0

2 years ago

You need to cross a canal and want to determine the distance across the opposite side. Since you're able to take measurements on

MariettaO [177]

Answer:

36.7 ft

Step-by-step explanation:

Measurements for the sides of the canal is given as: 40 ft. and 16 ft.

You solve the above question using Pythagoras Theorem

The distance across the canal is calculated as:

√(40² - 16²)

= √(1344)

= 36.66060556 ft

Approximately = 36.7 ft

Therefore, the distance x across the canal = 36.7 ft

8 0

2 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

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

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

2 years ago