All the letters in the word

Mathematics

2 answers:

Firlakuza [10]2 years ago

7 0

2/7 is the probability

kupik [55]2 years ago

5 0

Answer:

The probability of picking this is 2/7

Step-by-step explanation:

I'm pretty sure this is correct, and my reasoning for this is because there is 1 of most of the letters. This my not be correct, so you should probably try asking someone else, too :D!

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Need help with this question. Will crown brainliest

Fynjy0 [20]

Answer f(x)=1/x
g(x)=x+2
f(g(x))=f(x+2)
f(x+2)=1/(x+2)=h(x)

4 0

2 years ago

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 $\mathsf{sin\,\beta:}$

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

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

6 0

3 years ago

Pre-calc, image attached

makvit [3.9K]

Answer:

$\lim_{x \to 7} \frac{\sqrt{x+2}-3 }{x-7} = \frac{1}{6}$