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3 years ago

Is y = 0.860x + 3.302 a linear function?

Mathematics

1 answer:

gulaghasi [49]3 years ago

8 0

Answer:

yes

Step-by-step explanation:

Yes. The given function has the form y = mx + b, which is that of a linear function. Here the slope, m, is 0.860, and the y-intercept is 3.302.

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Can someone please help me find the domain, range, intercepts, and asymptotes pf this function?

marishachu [46]

1) given function

y = - 2 ^ ( -x + 2) + 1

2) domain: domain is the set of the x-values for which the function is defined.

The exponential function is defined for all the real numbers, so the domain of the given function is all the real numbers.

3) x-intercept => y = 0

=> y = - 2 ^ ( -x + 2) + 1 = 0 => 2^ ( -x + 2) = 1

=> - x + 2 = 0 => x = 2

The x-intercept is x = 0

4) y-intercept => x = 0

=> y = - 2 ^ ( -x + 2) + 1= - 2 ^ ( 0 + 2) 1 = - (2)^(2) + 1 =- 4 + 1 = - 3

=> The y-intercept is - 3

5) limit when x -> negative infinite

Lim f(x) when x -> ∞ = - ∞

6) limit when x -> infinite

Lim f(x) when x - > infinite = 1

=> asymptote = y = 1

7) range is the set of values of the fucntion: y

Given that the function is strictly decreasing from -∞ to ∞, the range is from - ∞ to less than 1

Range (-∞,1)

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3 years ago

Which description represents this equation?

ratelena [41]

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3 years ago

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anzhelika [568]

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3 years ago

What is .04 as a mixed number

statuscvo [17]

Step 1: 0.4 = 4⁄10
Step 2: Simplify 4⁄10 = 2⁄5

3 0

3 years ago

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.

Oliga [24]

Answer:

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

$\rm \displaystyle 2)3021.6 \: m$

Step-by-step explanation:

<h3>Question-1:</h3>

so when flash down 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 48.2 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 3021.6 meters high above sea-level the rocket gets at its peak?

5 0

3 years ago