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nasty-shy [4]
3 years ago
9

WILL MARK BRANLIESTIn a multiplication problem, if the first factor has 5 digits after the decimal and the second factor has 4 d

igits after the decimal, then the product will have _____ digits after the decimal.
1
4
5
9
Mathematics
2 answers:
victus00 [196]3 years ago
8 0
You can just pick any 5 digit decimal and multiply it with any 4 digit decimal to check.

You will see that it has 9 digits after the decimal.
laiz [17]3 years ago
8 0
The product of a 5 decimal-digit number by a 4-decimal digit number will have 8 digits after the decimal. For example:
0.12345 × 0.1234 = <span>0.01523373 </span>←8 digits
<span>However, we are multiplying with a whole number and a decimal, we will get 9 digits. Therefore, the answer to your query is 9. Hope this helps and have a great day!</span>
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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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