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postnew [5]
3 years ago
13

If the statement "If I am hungry, then I am not happy" is assumed to be true, is its inverse, "If I am not hungry, then I must b

e happy," also always true? A. No B. Yes
Mathematics
2 answers:
irga5000 [103]3 years ago
8 0
Hey there!

This is to be a true statement, each thing that he say's is the opposite.

(I am hungry) = I am not hungy

(I am not happy) = I am happy

They both are inversed

Your correct answer is . . . 

\boxed{\boxed{Yes, \ they \ are \ inversed}}

Hope this helps.
~Jurgen
Sophie [7]3 years ago
4 0
The answer will be B.yes plz mark me brainliest
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3 years ago
Please help!! 10 point available:)
salantis [7]

Answer:

1) - 2n + 14

2)- 5n + 30

Step-by-step explanation:

n = a + (n - 1) \times d \\  = 12 + (n - 1) \times  - 2 \\  = 12  - 2n + 2 \\  =  - 2n + 14

n = a + (n - 1) \times d \\  = 25 + (n - 1) \times  - 5 \\  = 25 - 5n + 5 \\  =  - 5n + 30

<em>Can</em><em> </em><em>I</em><em> </em><em>have</em><em> </em><em>the</em><em> </em><em>brainliest</em><em> </em><em>please</em><em>?</em>

8 0
3 years ago
Help me with trigonometry
poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the<em> </em><u><em>Weierstrass substitution</em></u>, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}

Therefore,

\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}

Once the double angle identity for sine is

\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}

we know \sin(x)=\dfrac{2t}{1+t^2}, but sure,  we can derive this formula considering the double angle identity

\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)

Recall

\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}

Thus,

\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}

Similarly for cosine, consider the double angle identity

Thus,

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

Hence, we showed \sin(x) \text { and } \cos(x)

======================================================

5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]

Solving

5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3

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t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} =  \dfrac{3\pm \sqrt{10}}{-2}

t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}

Just note that

\tan\left(\dfrac{x}{2}\right) =  \dfrac{3\pm 8\sqrt{10}}{-2}

and  \tan\left(\dfrac{x}{2}\right) is not defined for x=k\pi , k\in\mathbb{Z}

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5 0
3 years ago
Read 2 more answers
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