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3 years ago

7

The equation of line h is y = -2x + 1. If line h is dilated with a scale factor of 4 with respect to the origin, what is the equ

ation of the new line?

1 answer:

amm18123 years ago

3 0

Answer:

y = -2x + 4

Step-by-step explanation:

<u>Given line h:</u>

y = -2x + 1

Dilated line is parallel to original, so will have same slope of -2

<u>y-intercept will change by a scale factor of 4: </u>

(0, 1) → (0*4, 1*4) = (0, 4)

<u>So the equation of the new line is:</u>

y = -2x + 4

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Read 2 more answers

Generate the nest three terms of each arithmetic sequence shown below.

o-na [289]

Answer:

A)2,6,10

B)2,-6,-18

C)-1,-3,-5

Step-by-step explanation:

<u>A)a1=-2 and d=4</u>

We know that the arithmetic sequence formula is

$a_{n}=a_{1}+(n-1)d$

Now

$a_{2}=a_{1}+(2-1)d$

$a_{2}=a_{1}+(1)d$

Substituting the given value we get

$a_{2}= -2+(1)4$

$a_{2}= -2+4$

$a_{2}= 2$

------------------------------------------

Similarly

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

$a_{3}=a_{1}+(2)d$

Substituting the given value we get

$a_{3}= -2+(2)4$

$a_{3}= -2+8$

$a_{3}= 6$

-------------------------------------------------

$a_{4}=a_{1}+(4-1)d$

$a_{4}=a_{1}+(3)d$

Substituting the given value we get

$a_{4}= -2+(3)4$

$a_{4}= -2+16$

$a_{4}= 10$

-------------------------------------------------------------------------------------------

<u>B</u><u> $a_n=a_{(n-1)}-8$ with $a_1=10$ </u>

$a_2=a_{(2-1)}-8$

$a_2=a_{1}-8$

Substituting the given value

$a_2= 10-8$

$a_2=2$

---------------------------------------------------------------------

$a_3=a_{(3-1)}-8$

$a_3=a_{2}-8$

Substituting the value

$a_3=2-8$

$a_3= -6$

---------------------------------------------------------------------

$a_4=a_{(4-1)}-8$

$a_4=a_{3}-8$

Substituting the value

$a_4= -6-8$

$a_4= -14$

-------------------------------------------------------------------------------------------

<u>C) $a_1=3, a_2=1$ </u>

Here the difference is -2

the arithmetic sequence formula is

$a_{n}=a_{1}+(n-1)d$

Now

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

$a_{3}=a_{1}+(2)d$

Substituting the value we get

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

$a_{3}= 3-4$

$a_{3}= -1$

------------------------------------------------------------------------------

$a_{4}=a_{1}+(4-1)d$

$a_{4}=a_{1}+(3)d$

Substituting the value we get

$a_{4}= 3+(3)-2$

$a_{4}= 3-6$

$a_{4}= -3$

----------------------------------------------------------------------------------

$a_{5}=a_{1}+(5-1)d$

$a_{5}=a_{1}+(4)d$

Substituting the value we get

$a_{5}= 3+(4)-2$

$a_{5}= 3-8$

$a_{5}= -5$

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