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Korolek [52]
3 years ago
6

a video store charges a one-time membership fee of $12.00 plus $1.50 per video rental. How many videos can stewart rent if he sp

ends $21?
Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
4 0
He could rent 6 videos if he spends $21.00
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DochEvi [55]

Answer:

BOI YOU TALKING ABOUT SOME THIS LOOKS EASY, THIS DOESN'T EVEN LOOK EASY PLUS... i dont know the answer. Bye

Step-by-step explanation:

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What is the difference between a sequence and a series? What is the difference between a geometric and an arithmetic sequence?
Neporo4naja [7]

Answer/Step-by-step explanation:

A series is events coming one after another but a sequence is in a certain order.

arithmetic is the difference between two constant terms while geometric is the ratio between two constant terms.

arithmetic: -

geometric: ÷

5 0
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Without solving, identify whether each of the following equations has a unique solution, no solution, or infinitely many solutio
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No solution

Step-by-step explanation:

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HELP PLEASE!! 100 POINTS!! <br> What is the number of terms in this expression?<br><br> m/5+4⋅6
Tpy6a [65]

Answer:

m/5 is one term and it is separated by 4*6 by add sign there are two terms

Step-by-step explanation:

hope this helps.

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3 years ago
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Write the given expression in terms of x and y only.<br> sin(sin−1(x) + cos−1(y))
yan [13]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2308127

_______________


Write the expression below in terms of x and y only:

(I'm going to call it "E")

\mathsf{E=sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]\qquad\quad(i)}


Let

\begin{array}{lcl} \mathsf{\alpha=sin^{-1}(x)}&\qquad&\mathsf{then~-\,\dfrac{\pi}{2}\le \alpha\le \dfrac{\pi}{2}}\\\\\\ \mathsf{\beta=cos^{-1}(x)}&\qquad&\mathsf{then~0\le \beta\le \pi.} \end{array}


so the expression becomes

\mathsf{E=sin(\alpha+\beta)}\\\\ \mathsf{E=sin\,\alpha\,cos\,\beta+sin\,\beta\,cos\,\alpha\qquad\quad(ii)}


•   Finding \mathsf{sin\,\alpha:}

\mathsf{sin\,\alpha=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\alpha=x\qquad\quad\checkmark}


•   Finding \mathsf{cos\,\alpha:}

\mathsf{sin^2\,\alpha=x^2}\\\\ \mathsf{1-cos^2\,\alpha=x^2}\\\\ \mathsf{cos^2\,\alpha=1-x^2}\\\\ \mathsf{cos\,\alpha=\sqrt{1-x^2}\qquad\quad\checkmark}


because \mathsf{cos\,\alpha} is positive for \mathsf{\alpha\in \left[-\frac{\pi}{2},\,\frac{\pi}{2}\right].}


•   Finding \mathsf{cos\,\beta:}

\mathsf{cos\,\beta=cos\!\left[cos^{-1}(y)\right]}\\\\ \mathsf{cos\,\beta=y\qquad\quad\checkmark}


•   Finding \mathsf{sin\,\beta:}

\mathsf{cos^2\,\alpha=y^2}\\\\&#10; \mathsf{1-sin^2\,\beta=y^2}\\\\ \mathsf{sin^2\,\beta=1-y^2}\\\\ &#10;\mathsf{sin\,\beta=\sqrt{1-y^2}\qquad\quad\checkmark}


because \mathsf{sin\,\beta} is positive for \mathsf{\beta\in [0,\,\pi].}


Finally, you get

\mathsf{E=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}}\\\\\\ \therefore~~\mathsf{sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}\qquad\quad\checkmark}


I hope this helps. =)


Tags:   <em>inverse trigonometric trig function sine cosine sin cos arcsin arccos sum angles trigonometry</em>

6 0
3 years ago
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