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3 years ago

Any kind soul willing to help a random person lol

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

7 0

Um i got 6,42 rounded

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Use repeated subtraction to divide 35/4

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Find the exact value of cos(sin^-1(-5/13))

son4ous [18]

bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.

$\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}$

$\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}$

le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

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3 years ago

3(3a+2)+6=30 what does a equal

Naya [18.7K]

Answer:

a=2

Step-by-step explanation:

3(3a+2)+6=30

9a+6+6=30

9a+12=30

9a=18

a=2

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3 years ago

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What times what equals -30 but also can be added together and equal -4

Step2247 [10]

Answer:

Step-by-step explanation:

xy=-30

y=-30/x

x+y=-4

x-30/x=-4

x²-30=-4x

x²+4x-30=0

$x^2+4x-30=0\\x=\frac{-4\pm\sqrt{16+120} }{2} \\=\frac{-4\pm\sqrt{136} }{2} \\=-2\pm\sqrt{39} \\=-2+\sqrt{39} ,-2-\sqrt{39} \\y=\frac{-4}{-2+\sqrt{39} } ,\frac{4}{2+\sqrt{39} }$ $correction write ~-30 ~in~place~of~-4~in~the~end$

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3 years ago