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Llana [10]
2 years ago
12

What is the sign of 37 + (-37)?

Mathematics
2 answers:
Aleks04 [339]2 years ago
8 0

Answer:

<u>The result is zero:</u>

  • 37 + (-37) = 0

The sign is neither positive nor negative

Fantom [35]2 years ago
6 0

Hey there!

If we solve the following math problem, we'll get

37-37=0

0 is neither positive nor negative.

Hope it helps you!

GraceRosalia

Brainliest please! :)

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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>

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As the weather changes from Summer to Fall, it was noted that the temperature dropped 20°F over 4 weeks. It continued to drop th
mihalych1998 [28]

Answer:

The integer that represents the change in the temperature each week = 5

The temperature of the weather drops 5°F each week

Step-by-step explanation:

The given parameters regarding the weather are;

The range over which the temperature dropped = 20°F

The duration over which the temperature dropped = 4 weeks

The rate at which the weather dropped each week = The same amount (constant rate)

Therefore, we have;

For a constant rate of change, the rate of change of the temperature of the weather, m, is given as follows;

m = (The change in the temperature)/(The duration of the change)

m = 20°F/(4 weeks) = 5°F/week

The integer that represents the change in the temperature each week = 5

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3 years ago
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