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3 years ago

Y=4x+3 table of ordered pairs

Mathematics

1 answer:

lara31 [8.8K]3 years ago

6 0

Answer:

X:1 2 3 4 5 Y:7 11 15 19 23.

Step-by-step explanation:

We just need to put in the values of x into the equation: y = 4x + 3.

x = 1:

y = 4(1) + 3

= 4 + 3

y = 7

x = 2:

y = 4(2) + 3

= 8 + 3

y = 11

x = 3:

y = 4(3) + 3

= 12 + 3

y = 15

x = 4:

y = 4(4) + 3

= 16 + 3

y = 19

x = 5:

y = 4(5) + 3

= 20 + 3

y = 23

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Setler79 [48]

Answer:

(a) The recurrence relation for the salary is

$S_{n+1}=1.05*S_n+1000\\\\S_0=50000$

(b) The salary 25 years after 2017 will be $217044.85.

Step-by-step explanation:

We can define the next year salary $S_{n+1}$ as

$S_{n+1}=S_n+1000+0.05*S_n=1.05*S_n+1000$

wit S0=$50000

If we extend this to 2 years from 2017 (n+2), we have

$S_{n+2}=1.05*S_{n+1}+1000=1.05*(1.05*S_n+1000)+1000\\S_{n+2} =1.05^2*S_n+1.05*1000+1000\\S_{n+2}=1.05^2*S_n+1000*(1.05^1+1)$

Extending to 3 years (n+3)

$S_{n+3}=1.05*S_{n+2}+1000=1.05(1.05^2*S_n+1000*(1.05^1+1))+1000\\\\S_{n+3}=1.05^3S_n+1.05*1000*(1.05^1+1)+1000\\\\S_{n+3}=1.05^3*S_n+1000*(1.05^2+1.05^1+1)$

Extending to 4 years (n+4)

$S_{n+4}=1.05*S_{n+3}+1000=1.05*(1.05^3*S_n+1000*(1.05^2+1.05^1+1))+1000\\\\S_{n+4}=1.05^4S_n+1.05*1000*(1.05^2+1.05^1+1))+1000\\\\S_{n+4}=1.05^4S_n+1000*(1.05^3+1.05^2+1.05^1+1.05^0)$

We can now express a general equation for S_n (salary at n years from 2017)

$S_n=1.05^nS_0+1000*\sum_{0}^{n-1}1.05^n$

The salary at 25 years from 2017 (n=25) will be

$S_{25}=1.05^{25}S_0+1000*\sum_{0}^{24}1.05^i\\\\S_{25}=3.386*50000+1000*47.72=217044.85$

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4 years ago

4. If f(x) = 4x + 1 and g(x) = 2x2, what is the value of g(f(-8))?

ASHA 777 [7]

Answer:

A. 1922

Step-by-step explanation:

f(x) = 4x + 1 and g(x) = 2x^2

First find f (-8)

f(-8) = 4(-8) + 1 = -32+1 = -31

Then find g(-31)

g(-31) = 2 (-31)^2 =1922

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