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wolverine [178]
3 years ago
9

(-1,-2) and (3, -162) a. Is there a line th(at goes through these two points? How do you know? If there is one, what is the equa

tion of the line going through these two points? if there isn't one, explain why not.
Mathematics
1 answer:
lana [24]3 years ago
4 0

Answer:

The equation of line for given points is y = - 40 x - 42  

Step-by-step explanation:

Given two points as :

( -1 , - 2 )  and ( 3 , - 162 )

Now, Equation of line in points form

y - y_1 = m × ( x -  x_1 )

Where m is the slope of line

And, m = \frac{y_2 - y_1}{x_2 - x_1}

or, m = \frac{ - 162 + 2}{ 3 + 1}

or, m =  \frac{ - 160}{ 4}

∴   m = - 40

So, Slope of line for given points is - 40

Now equation of line can be written as

y - ( - 2 ) = ( - 40 ) × ( x -  ( - 1 ) )

or, y + 2 =  ( - 40 ) × ( x + 1 )

or, y + 2 = - 40 x - 40

or, y = - 40 x - 40 - 2

∴  y = - 40 x - 42        

Hence the equation of line for given points is y = - 40 x - 42    Answer

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NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h ( t ) = − 4.
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Answer:

\displaystyle 1)48.2    \:  \: \text{sec}

\rm \displaystyle  2)3021.6 \: m

Step-by-step explanation:

<h3>Question-1:</h3>

so when <u>flash down</u><u> </u>occurs the rocket will be in the ground in other words the elevation(height) from ground level will be 0 therefore,

to figure out the time of flash down we can set h(t) to 0 by doing so we obtain:

\displaystyle  - 4.9 {t}^{2}  + 229t + 346 = 0

to solve the equation can consider the quadratic formula given by

\displaystyle x =  \frac{ - b \pm  \sqrt{ {b}^{2} - 4 ac} }{2a}

so let our a,b and c be -4.9,229 and 346 Thus substitute:

\rm\displaystyle t =  \frac{ - (229) \pm  \sqrt{ {229}^{2} - 4.( - 4.9)(346)} }{2.( - 4.9)}

remove parentheses:

\rm\displaystyle t =  \frac{ - 229 \pm  \sqrt{ {229}^{2} - 4.( - 4.9)(346)} }{2.( - 4.9)}

simplify square:

\rm\displaystyle t =  \frac{ - 229 \pm  \sqrt{ 52441- 4( - 4.9)(346)} }{2.( - 4.9)}

simplify multiplication:

\rm\displaystyle t =  \frac{ - 229 \pm  \sqrt{ 52441- 6781.6} }{ - 9.8}

simplify Substraction:

\rm\displaystyle t =  \frac{ - 229 \pm  \sqrt{ 45659.4} }{ - 9.8}

by simplifying we acquire:

\displaystyle t = 48.2  \:  \:  \: \text{and} \quad  - 1.5

since time can't be negative

\displaystyle t = 48.2

hence,

at <u>4</u><u>8</u><u>.</u><u>2</u><u> </u>seconds splashdown occurs

<h3>Question-2:</h3>

to figure out the maximum height we have to figure out the maximum Time first in that case the following formula can be considered

\displaystyle x _{  \text{max}} =  \frac{ - b}{2a}

let a and b be -4.9 and 229 respectively thus substitute:

\displaystyle t _{  \text{max}} =  \frac{ - 229}{2( - 4.9)}

simplify which yields:

\displaystyle t _{  \text{max}} =  23.4

now plug in the maximum t to the function:

\rm \displaystyle  h(23.4)- 4.9 {(23.4)}^{2}  + 229(23.4)+ 346

simplify:

\rm \displaystyle  h(23.4)  =  3021.6

hence,

about <u>3</u><u>0</u><u>2</u><u>1</u><u>.</u><u>6</u><u> </u>meters high above sea-level the rocket gets at its peak?

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