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zhannawk [14.2K]

3 years ago

5

The set of ordered pairs shown represents a function f. { (4, -2), (-1, -1), (0, 0), (1, -1), (4,2) } Select the ordered pair th

at could be added to the set so that f remains a function.
A) (0, 3)
B) (9, -3)
C) (1, - 2)
D) (4, - 4)

1 answer:

Blababa [14]3 years ago

3 0

I will be right back to help with

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By using double inequality describe all the numbers that are:

atroni [7]

Question is -5 and -3

8 0

3 years ago

How to do this & what the answer is PLSSSSS HELP

Contact [7]

Answer:

<em>We</em><em> </em><em>can</em><em> </em><em>say</em><em> </em><em>that</em>

<em> </em>3x + x + 8 = 32

<em>So</em><em>:</em><em> </em>

3x + x + 8 = 32

4x = 32 - 8

4x = 24

x = 24/4

<em><u>x</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>6</u></em>

6 0

3 years ago

Find the third order maclaurin polynomial. Use it to estimate the value of sqrt1.3

vodka [1.7K]

$\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}$ is the maclaurin polynomial and estimate value of $\sqrt{1.3}$ is 1.14. This can be obtained by using the formula to find the maclaurin polynomial.

<h3>Find the third order maclaurin polynomial:</h3>

Given the polynomial,

$f(x)=\sqrt{1+3x}=(1+3x)^{\frac{1}{2} }$

The formula to find the maclaurin polynomial,

$f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3}$

Next we have to find f'(x), f''(x) and f'''(x),

$f'(x) = \frac{3}{2}(1+3x)^{-\frac{1}{2} }$

$f''(x) =-\frac{9}{4}(1+3x)^{-\frac{3}{2} }$

$f'''(x) = \frac{81}{8}(1+3x)^{-\frac{5}{2} }$

By putting x = 0 , we get,

$f(0)=(1+3(0))^{\frac{1}{2} }=1$

$f'(0) = \frac{3}{2}(1+3(0))^{-\frac{1}{2} }=\frac{3}{2}$

$f''(0) =-\frac{9}{4}(1+3(0))^{-\frac{3}{2} }=-\frac{9}{4}$

$f'''(0) = \frac{81}{8}(1+3(0))^{-\frac{5}{2} }=\frac{81}{8}$

Therefore the maclaurin polynomial by using the formula will be,

$\sqrt{1+3x}=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3}$

$\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}$

To find the value of $\sqrt{1.3}$ we can use the maclaurin polynomial,

$\sqrt{1.3}$ is $\sqrt{1+3x}$ with x = 1/10,

$\sqrt{1+3(1/10)}=1+\frac{3}{2} (1/10)-\frac{9}{8} (1/10)^{2} + \frac{81}{8}(1/10)^{3}$

$\sqrt{1+3(1/10)}=\frac{18247}{16000} = 1.14$

Hence $\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}$ is the maclaurin polynomial and estimate value of $\sqrt{1.3}$ is 1.14.

Learn more about maclaurin polynomial here:

brainly.com/question/24188694

#SPJ1

6 0

2 years ago

Read 2 more answers

7 Given: -8(w + 1) = -5(w + 10); Prove: w = 14<br> Statements

LenKa [72]

Answer:

w=14

Step-by-step explanation:

-8(w + 1) = -5(w + 10)

Remove the parentheses

-8w-8+-5w-50

Move terms

-8w+5w=-50+8

Collect like terms and calculate

-3w=-42

Then divide on both sides

5 0

3 years ago

Can some one help me in this problem

strojnjashka [21]

I believe x=116 and z=26. 4z-40=64
4z=104
z=26
180-64=116

8 0

3 years ago