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Nostrana [21]
4 years ago
13

Idk, I need help...........

Mathematics
1 answer:
elena55 [62]4 years ago
4 0

Answer:

2

Step-by-step explanation:

as the 0.8 is between 1 and 0 (that all are on the x axis) and given in graph that 2, (that is on the y axis), connects the dots between the 0 & 1 so yea

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Find the exact value of tan (arcsin (two fifths)). For full credit, explain your reasoning.
Hitman42 [59]
\bf sin^{-1}(some\ value)=\theta \impliedby \textit{this simply means}
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sin(\theta )=some\ value\qquad \textit{now, also bear in mind that}
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sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad 
\qquad 
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\

\bf sin^{-1}\left( \frac{2}{5} \right)=\theta \impliedby \textit{this simply means that}
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sin(\theta )=\cfrac{2}{5}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\qquad \textit{now let's find the adjacent side}
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\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}

\bf \pm \sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a
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\textit{we don't know if it's +/-, so we'll assume is the + one}\quad \sqrt{21}=a\\\\
-------------------------------\\\\
tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{2}{\sqrt{21}}
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\textit{and now, let's rationalize the denominator}
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\cfrac{2}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{(\sqrt{21})^2}\implies \cfrac{2\sqrt{21}}{21}

7 0
3 years ago
If a = √3-√11 and b = 1 /a, then find a² - b²​
Serga [27]

If b=\frac1a, then by rationalizing the denominator we can rewrite

b = \dfrac1{\sqrt3-\sqrt{11}} \times \dfrac{\sqrt3+\sqrt{11}}{\sqrt3+\sqrt{11}} = \dfrac{\sqrt3+\sqrt{11}}{\left(\sqrt3\right)^2-\left(\sqrt{11}\right)^2} = -\dfrac{\sqrt3+\sqrt{11}}8

Now,

a^2 - b^2 = (a-b) (a+b)

and

a - b = \sqrt3 - \sqrt{11} + \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{9\sqrt3 - 7\sqrt{11}}8

a + b = \sqrt3 - \sqrt{11} - \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{7\sqrt3 - 9\sqrt{11}}8

\implies a^2 - b^2 = \dfrac{\left(9\sqrt3 - 7\sqrt{11}\right) \left(7\sqrt3 - 9\sqrt{11}\right)}{64} = \boxed{\dfrac{441 - 65\sqrt{33}}{32}}

5 0
2 years ago
You have two cards with a sum of (-12) in your hands. what two cards could you have?
Readme [11.4K]
Your two cards could be a negative six and a positive two.
4 0
3 years ago
C+4c=15 what’s the answer?
Fynjy0 [20]
C=3 that’s the answer I think
4 0
3 years ago
If f(3) = 9, what is f(1)<br>​
Vaselesa [24]

Answer:

3

Step-by-step explanation:

3*3=9

1*3=3

8 0
3 years ago
Read 2 more answers
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