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2 years ago

How do you represent a table with linear functions

1 answer:

8 0

A table can be represented with a linear function equation as y = mx + b, where m is the slope and b is the y-intercept.

<h3>How to Represent a Table with Linear Function?</h3>

Assuming we have a table of values as shown in the image attached below, to write an equation of linear function for the table, do the following:

Pick two pairs of values, say, (1, 5) and (2, 25) and find the slope (m):

Slope (m) = change in y / change in x = (25 - 5)/(2 - 1)

Slope (m) = 20

Find the y-intercept (b) by substituting (x, y) = (1, 5) and m = 20 into y = mx + b:

5 = 20(1) + b

5 = 20 + b

5 - 20 = b

-15 = b

b = -15

Write the equation of the linear function by substituting m = 20 and b = -15 into y = mx + b:

y = 20x - 15

Learn more about the linear function on:

brainly.com/question/15602982

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defon

Hello,

14/25 = x/275

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x = 275 * 14 / 25 = 154

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3 years ago

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A repairman charged $93.06. The price included 2 hours of labor and a $40 service charge.

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3 years ago

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Suppose it is known that for a given differentiable function y=f(x), its tangent line (local linearization) at the point where a

AleksandrR [38]

Answer:

y(-4) = 5

y'(-4) = -7

Step-by-step explanation:

Hi!

Since the tangent line T and the curve y must coincide at x=-4

y(-4) = T(-4) = 5

On the other hand, the derivative of the curve evaluated at -4 y'(x=-4) must be the slope of the tangent line. Which inspecting the tangent line T(x) is -7

That is:

y'(-4) = -7

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3 years ago

Please Help!! Emergency!! I will make you brainiest and give you 15 points!!!!!

bonufazy [111]

Answer:

see explanation

Step-by-step explanation:

Under a rotation about the origin of 90°

a point (x, y ) → (- y, x ), thus

A(2, 2 ) → A'(- 2, 2 )

B(2, 4 ) → B'(- 4, 2 )

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D(6, 4 ) → D'(- 4, 6 )

E(6, 2 ) → E'(- 2, 6 )

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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 <em>0.002652</em> 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

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