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daser333 [38]
2 years ago
12

Which is NOT the graph of a function?

Mathematics
1 answer:
BabaBlast [244]2 years ago
4 0

Answer:

B. it's not useful for this question

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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.
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 <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?

5 0
2 years ago
Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^
Gemiola [76]

Answer:

\rm \displaystyle y' =   2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x}

Step-by-step explanation:

we would like to figure out the differential coefficient of e^{2x}(1+\ln(x))

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

\displaystyle y =  {e}^{2x}  \cdot (1 +   \ln(x) )

to do so distribute:

\displaystyle y =  {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x}

take derivative in both sides which yields:

\displaystyle y' =  \frac{d}{dx} ( {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x} )

by sum derivation rule we acquire:

\rm \displaystyle y' =  \frac{d}{dx}  {e}^{2x}  +  \frac{d}{dx}   \ln(x)  \cdot  {e}^{2x}

Part-A: differentiating $e^{2x}$

\displaystyle \frac{d}{dx}  {e}^{2x}

the rule of composite function derivation is given by:

\rm\displaystyle  \frac{d}{dx} f(g(x)) =  \frac{d}{dg} f(g(x)) \times  \frac{d}{dx} g(x)

so let g(x) [2x] be u and transform it:

\displaystyle \frac{d}{du}  {e}^{u}  \cdot \frac{d}{dx} 2x

differentiate:

\displaystyle   {e}^{u}  \cdot 2

substitute back:

\displaystyle    \boxed{2{e}^{2x}  }

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

\displaystyle  \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)

let

  • f(x) \implies   \ln(x)
  • g(x) \implies    {e}^{2x}

substitute

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =  \frac{d}{dx}( \ln(x) ) {e}^{2x}  +  \ln(x) \frac{d}{dx}  {e}^{2x}

differentiate:

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =   \boxed{\frac{1}{x} {e}^{2x}  +  2\ln(x)  {e}^{2x} }

Final part:

substitute what we got:

\rm \displaystyle y' =   \boxed{2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x} }

and we're done!

6 0
3 years ago
Add: 3x(x-y+z) and 2y (x+y-z)​
hodyreva [135]

Answer:

3x^2 + 3xz + 2y^2 - 2yz - xy

Step-by-step explanation:

(1) multiply it to get

3x^2 - 3xy + 3xz and 2yx + 2y^2 - 2yz

add similar variables (-3xy + 2yx)

( the order of the variables doesn't matter xy or yx is the same)

we get -xy

and then after that simplifying we get 3x^2 + 3xz + 2y^2 - 2yz - xy

3 0
3 years ago
Read 2 more answers
Independent random samples are selected from two populations and are used to test the hypothesis H0:(μ1−μ2)=0H0:(μ1−μ2)=0 agains
lozanna [386]

Answer:

a. We do not have enough evidence to conclude that the two means differ.

b. There is no difference in the population means.

Step-by-step explanation:

a) Results of the test:

Having p - value to be very high, we have strong evidence of not rejecting H o and hence,

We do not have enough evidence to conclude that the two means differ.

b) For a one tailed test,

p - value = 0.115/2 = 0.0575

Interpretation: p - value is the probability of getting the difference as small as obtained in the sample, given that the null hypothesis is true i.e. there is no difference in the population means.

7 0
3 years ago
Please help on a online test and WILL FAIL if I don’t get this right Find the missing factor.
Neko [114]

Answer:

(x+7)

Step-by-step explanation:

(x+7)(2x+1)=2x^2+14x+x+7=2x^2+15x+7

5 0
3 years ago
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