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KIM [24]
4 years ago
7

Where do i label -8/7?

Mathematics
1 answer:
Dmitriy789 [7]4 years ago
5 0

Answer:

Between minus one and zero

Step-by-step explanation:

The answer is between minus one amd zero

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Simplify the expression 3(x + 4)
Varvara68 [4.7K]
3(x+ 4)
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6 0
3 years ago
Read 2 more answers
Does anyone know how to do this ??? <br> -view attachment
cricket20 [7]

As soon as I read this, the words "law of cosines" popped
into my head.  I don't have a good intuitive feeling for the
law of cosines, but I went and looked it up (you probably
could have done that), and I found that it's exactly what
you need for this problem.

The "law of cosines" relates the lengths of the sides of any
triangle to the cosine of one of its angles ... just what we need,
since we know all the sides, and we want to find one of the angles.

To find angle-B, the law of cosines says

       b² = a² + c² - 2 a c cosine(B)

B  =  angle-B
b  =  the side opposite angle-B = 1.4
a, c = the other 2 sides = 1 and 1.9

                  (1.4)² = (1)² + (1.9)² - (2 x 1 x 1.9) cos(B)

                 1.96  =  (1) + (3.61)  -  (3.8) cos(B)

Add  3.8 cos(B)  from each side:

                 1.96 + 3.8 cos(B) = 4.61

Subtract  1.96  from each side:

                             3.8 cos(B) =  2.65

Divide each side by  3.8 :

                                  cos(B)  =  0.69737  (rounded)

Whipping out the
trusty calculator:
                                 B  = the angle whose cosine is 0.69737

                                      =  45.784° .

Now, for the first time, I'll take a deep breath, then hold it
while I look back at the question and see whether this is
anywhere near one of the choices ...

By gosh !  Choice 'B' is  45.8° !                    yay !
I'll bet that's it !

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Write the expression i35 in the standard form
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Find the exact value of the trigonometric expression.
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\bf \textit{Half-Angle Identities}&#10;\\\\&#10;sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}}&#10;\qquad &#10;cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}}\\\\&#10;-------------------------------\\\\&#10;\cfrac{1}{12}\cdot 2\implies \cfrac{1}{6}\qquad therefore\qquad \cfrac{\pi }{12}\cdot 2\implies \cfrac{\pi }{6}~thus~ \cfrac{\quad \frac{\pi }{6}\quad }{2}\implies \cfrac{\pi }{12}

\bf sec\left( \frac{\pi }{12} \right)\implies sec\left( \cfrac{\frac{\pi }{6}}{2} \right)\implies \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\impliedby \textit{now, let's do the bottom}&#10;\\\\\\&#10;cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{1+cos\left( \frac{\pi }{6} \right)}{2}}\implies \pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies \pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}}

\bf \pm \sqrt{\cfrac{2+\sqrt{3}}{4}}\implies \pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\implies \pm \cfrac{\sqrt{2+\sqrt{3}}}{2}&#10;\\\\\\&#10;therefore\qquad \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\implies \cfrac{2}{\sqrt{2+\sqrt{3}}}


which simplifies thus far to 

\bf \cfrac{2}{\sqrt{2+\sqrt{3}}}\cdot \cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\cdot \stackrel{conjugate}{\cfrac{2-\sqrt{3}}{2-\sqrt{3}}}&#10;\\\\\\&#10;\cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{2^2-(\sqrt{3})^2}\implies \cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{1}

\bf 2\sqrt{2+\sqrt{3}}(2-\sqrt{3})\impliedby \stackrel{\textit{keep in mind that}}{(2-\sqrt{3})=(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}})}&#10;\\\\\\&#10;2\sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}&#10;\\\\\\&#10;2\sqrt{(2+\sqrt{3})(2-\sqrt{3})\cdot (2-\sqrt{3})}&#10;\\\\\\&#10;2\sqrt{[2^2-(\sqrt{3})^2]\cdot (2-\sqrt{3})}\implies 2\sqrt{1(2-\sqrt{3})}&#10;\\\\\\&#10;2\sqrt{2-\sqrt{3}}
7 0
4 years ago
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