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3 years ago

5

A line has y intercept -6 and a x-intercept - 12. what is the equation of the line

1 answer:

koban [17]3 years ago

5 0

Answer:

$\boxed{y = -\dfrac{1}{2}x-6}}$

Step-by-step explanation:

The equation for a straight line is

y = mx + b

where m is the slope of the line and b is the y-intercept.

The line passes through the points (-12, 0) and (0, 6)

(a) Calculate the slope of the line

$\begin{array}{rcl}m & =&\dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\ & = &\dfrac{6 - 0 }{0 - (-12)}\\\\& = &\dfrac{6}{-12}\\\\& = & -\dfrac{1}{2}\\\end{array}$

(b) Write the equation

The y-intercept is at x = -6.

$\text{The equation for the line is $\boxed{\mathbf{y = -\dfrac{1}{2}x-6}}$}$

The diagram shows the graph of the line passing through the two intercepts.

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<h3>Solution:</h3>

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Step-by-step explanation:

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3 years ago

Consider the sequence defined recursively by do = 0, an = an-1 + 3n – 1. a) Write out the first 5 terms of this sequence,

creativ13 [48]

Answer:

The first 5 terms of the sequence is 2,7,15,26,40.

Step-by-step explanation:

Given : Consider the sequence defined recursively by $a_0=0$ $a_n=a_{n-1}+3n-1$

To find : Write out the first 5 terms of this sequence ?

Solution :

$a_n=a_{n-1}+3n-1$ and $a_0=0$

The first five terms in the sequence is at n=1,2,3,4,5

For n=1,

$a_1=a_{1-1}+3(1)-1$

$a_1=a_{0}+3-1$

$a_1=0+2$

$a_1=2$

For n=2,

$a_2=a_{2-1}+3(2)-1$

$a_2=a_{1}+6-1$

$a_2=2+5$

$a_2=7$

For n=3,

$a_3=a_{3-1}+3(3)-1$

$a_3=a_{2}+9-1$

$a_3=7+8$

$a_3=15$

For n=4,

$a_4=a_{4-1}+3(4)-1$

$a_4=a_{3}+12-1$

$a_4=15+11$

$a_4=26$

For n=5,

$a_5=a_{5-1}+3(5)-1$

$a_5=a_{4}+15-1$

$a_5=26+14$

$a_5=40$

The first 5 terms of the sequence is 2,7,15,26,40.

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Step-by-step explanation:

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