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Ronch [10]
3 years ago
15

Jane, Scott, and Frank have a total of $144 in their wallets. Frank has 4 times what Jane has. Scott has $6 less than Jane. How

much does each have?
Mathematics
1 answer:
konstantin123 [22]3 years ago
6 0

Answer:

Jane have $25

Frank have $100

Scott have $19

Step-by-step explanation:

To find the amount that each have, we have to represent this information in a mathematical format  and then solve

let x represent represent the amount that Jane has

let Y represent represent the amount that Frank has

let z represent represent the amount that Scott has

From the question given, "Jane, Scott, and Frank have a total of $144"  can be represented mathematically as

x + y + z = $144 ---------------------------------------------------------------------------------(1)

"Frank has 4 times what Jane has"   can be represented mathematically as

y = 4x ---------------------------------------------------------------------------------------------(2)

"Scott has $6 less than Jane"  can be represented mathematically as

z= x- 6 ----------------------------------------------------------------------------------------------(3)

we can now solve the equations using the substitution method

substitute equation (2)  and (3)  into equation (1)

x + y + z = $144

x + 4x + x-6 = $144

6x -6 = $144

add 6 to both-side of the equation

6x - 6 + 6 = $144 + 6

6x =$150

Divide both-side of the equation by 6

6x/6 = 150/6

x = $25

from equation 2

y = 4x

substitute x= $25 in equation (2)

y = 4 (25)

y = $100

substitute x=$25 in equation  (3)

z = x- 6

z = 25 - 6

z=$19

Jane have $25

Frank have $100

Scott have $19

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If you're using the app, try seeing this answer through your browser:  brainly.com/question/2264253

_______________


Evaluate the indefinite integral:

\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx\qquad\quad\checkmark}


Trigonometric substitution:

\mathsf{\theta=sin^{-1}(x)\qquad\qquad\dfrac{\pi}{2}\le \theta\le \dfrac{\pi}{2}}


then,

\begin{array}{lcl} \mathsf{x=sin\,\theta}&\quad\Rightarrow\quad&\mathsf{dx=cos\,\theta\,d\theta\qquad\checkmark}\\\\\\ &&\mathsf{x^2=sin^2\,\theta}\\\\ &&\mathsf{x^2=1-cos^2\,\theta}\\\\ &&\mathsf{cos^2\,\theta=1-x^2}\\\\ &&\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\checkmark}\\\\\\ &&\textsf{because }\mathsf{cos\,\theta}\textsf{ is positive for }\mathsf{\theta\in \left[\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right].} \end{array}


So the integral \mathsf{(ii)} becomes

\mathsf{=\displaystyle\int\! \theta\,cos\,\theta\,d\theta\qquad\quad(ii)}


Integrate \mathsf{(ii)} by parts:

\begin{array}{lcl} \mathsf{u=\theta}&\quad\Rightarrow\quad&\mathsf{du=d\theta}\\\\ \mathsf{dv=cos\,\theta\,d\theta}&\quad\Leftarrow\quad&\mathsf{v=sin\,\theta} \end{array}\\\\\\\\ \mathsf{\displaystyle\int\!u\,dv=u\cdot v-\int\!v\,du}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-\int\!sin\,\theta\,d\theta}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-(-cos\,\theta)+C}

\mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta+cos\,\theta+C}


Substitute back for the variable x, and you get

\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=sin^{-1}(x)\cdot x+\sqrt{1-x^2}+C}\\\\\\\\ \therefore~~\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=x\cdot\,sin^{-1}(x)+\sqrt{1-x^2}+C\qquad\quad\checkmark}


I hope this helps. =)


Tags:  <em>integral inverse sine function angle arcsin sine sin trigonometric trig substitution differential integral calculus</em>

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