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valkas [14]
3 years ago
11

Find the six trigonometric function values for angle ∅ where its adjacent side is -9 and its hypotenuse is 41. (Theta is located

above the 90° angle of the right triangle for reference).

Mathematics
1 answer:
arsen [322]3 years ago
3 0
Check the picture below.

\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\
-------------------------------

\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~  cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~  tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}
\\\\\\
csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}

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What is the probability that a point chosen at random in the given figure will be inside the larger square and outside the small
In-s [12.5K]

Answer:

P(inside larger square and outside smaller) = \frac{51}{100}

Step-by-step explanation:

Probability is the result of the division of the number of possible outcome by the number of an event.

In the question, for a point chosen, the point can be in the small square only or in the area or region between the small square and the big square as such,

Area of larger square = area of region between both squares + area of smaller square

Where the area of a square is S × S where S is the side of a square

Area of larger square = 10 × 10

                                    = 100 cm square

Area of smaller square = 7 × 7

                                      = 49 cm square

Area of the region between  both squares

                                      = 100 - 49

                                      = 51 cm square

The probability that a dot selected is inside the larger square and outside the smaller is

P(inside larger square and outside smaller) = Area of region between both square/ Area of larger square

P(inside larger square and outside smaller) = \frac{51}{100}

5 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx....%7D%20%7D%20%7D%20%7D%20%20%3D%20%
andrey2020 [161]

First observe that if a+b>0,

(a + b)^2 = a^2 + 2ab + b^2 \\\\ \implies a + b = \sqrt{a^2 + 2ab + b^2} = \sqrt{a^2 + ab + b(a + b)} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{\cdots}}}}

Let a=0 and b=x. It follows that

a+b = x = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}}

Now let b=1, so a^2+a=4x. Solving for a,

a^2 + a - 4x = 0 \implies a = \dfrac{-1 + \sqrt{1+16x}}2

which means

a+b = \dfrac{1 + \sqrt{1+16x}}2 = \sqrt{4x + \sqrt{4x + \sqrt{4x + \sqrt{\cdots}}}}

Now solve for x.

x = \dfrac{1 + \sqrt{1 + 16x}}2 \\\\ 2x = 1 + \sqrt{1 + 16x} \\\\ 2x - 1 = \sqrt{1 + 16x} \\\\ (2x-1)^2 = \left(\sqrt{1 + 16x}\right)^2

(note that we assume 2x-1\ge0)

4x^2 - 4x + 1 = 1 + 16x \\\\ 4x^2 - 20x = 0 \\\\ 4x (x - 5) = 0 \\\\ 4x = 0 \text{ or } x - 5 = 0 \\\\ \implies x = 0 \text{ or } \boxed{x = 5}

(we omit x=0 since 2\cdot0-1=-1\ge0 is not true)

3 0
1 year ago
Can someone help me with this ? im struggling ​
snow_lady [41]

Answer:

Can you state in question in words. I cant see the picture.

Step-by-step explanation:

7 0
3 years ago
Arithmetic??????????????????
wel

good lesson and good day ^-^

7 0
2 years ago
What is the sum when 1/5 is added to 5/2?
NeTakaya

Answer:

27/10

Step-by-step explanation:

6 0
3 years ago
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