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3 years ago

Please help me - 9th grade math

Mathematics

1 answer:

ki77a [65]3 years ago

6 0

1. Height and weight 2. As height increases weight increases

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Can someone please help me

Allisa [31]

Answer:

I think it is C

Step-by-step explanation:

Reason being because the question said at least 15 points more so it can't

be 12 equal to 12 points. so you could eliminate the greater than or equal to.

so that leaves c and d. can't see how it is D because if he has 15 points more can't see why his other portion of points could take away 15 so that leaves C.

hope that helps

4 0

3 years ago

What is the domain and range of f(x)=2x-5

Mrrafil [7]

f(x) is a linear function with no restrictions

Domain: x = All real numbers

Range: y = All real numbers

8 0

3 years ago

Help 10 minutes left!

Annette [7]

The answer is the first one

5 0

3 years ago

Evaluate the numerical expression 2.5(−4.4 − 3.5)

Sophie [7]

2.5(−4.4 − 3.5)
=−4.4 − 3.5=-7.9
=2.5(-7.9)
=-19.75

6 0

4 years ago

Write the given expression in terms of x and y only. 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 $\mathsf{sin\,\beta:}$

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

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: inverse trigonometric trig function sine cosine sin cos arcsin arccos sum angles trigonometry

6 0

4 years ago