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3 years ago

14

Identify the x-intercept and y-intercept of the equation below.

1 answer:

Lady_Fox [76]3 years ago

8 0

Divide by 21 to put the equation in intercept form.
x/(21/9) + y/(-21/7) = 1
x/(7/3) + y/(-3) = 1

The x-intercept is (7/3, 0)
The y-intercept is (0, -3)

The 3rd choice is appropriate.

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Does 6(x+2)>x-3 have a solution

r-ruslan [8.4K]

Hi there!

To see if this has a solution, we can simplify it.

6x + 12 > x - 3
5x + 12 > -3
5x > -15
x > -3

Yes, there are solutions to this problem. Any number greater than -3 would match the inequality requirements.

Hope this helps!

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3 years ago

Create a table to find the third differences for the polynomial s^3 − s^2 +s for integer values of ss from −3 to 3.

evablogger [386]

Step-by-step explanation:

$f(s)=s^3-s^2$

at s = -3

$f(-3)=-3^3-(-3)^2=18$

at s = -2

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at s = -1

$f(-1)=-1^3-(-1)^2\\\Rightarrow f(-1)=-2$

at s = 0

$f(0)=0^3-0^2\\\Rightarrow f(0)=0$

at s = 1

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at s = 2

$f(2)=2^3-2^2\\\Rightarrow f(2)=4$

at s = 3

$f(3)=3^3-3^2\\\Rightarrow f(3)=18$

First difference

$f(-2')=f(-2)-f(-3)=-12-18=-30$

$f(-1')=f(-1)-f(-2')=-2--12=10$

$f(0')=f(0)-f(-1)=0--2=2$

$f(1')=f(1)-f(0)=0-0=0$

$f(2')=f(2)-f(1)=4-0=4$

$f(3')=f(3)-f(2)=18-4=14$

Second difference

$f(-1'')=f(-1')-f(-2')=10--30=40$

$f(0'')=f(0')-f(-1')=-8-40=-48$

$f(1'')=f(1')-f(0')=-2--8=6$

$f(2'')=f(2')-f(1')=14-4=10$

Third difference

$f(0'')-f(-1'')=-8-40=48$

$f(1'')-f(0'')=-2--8=6$

$f(2'')-f(1'')=4--2=6$

$f(2'')-f(1'')=10-4=6$

18 -12 -2 0 0 4 18

-30 10 2 0 4 14

40 -8 -2 4 10

-48 6 6 6

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