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3 years ago

HELP NO LINKS NO FILES JUST ANSWERS DUE TODAY

Mathematics

1 answer:

yawa3891 [41]3 years ago

4 0

Answer: From Figure A to Figure B it’s 1/2

Step-by-step explanation:

Count the squares around each triangle and then divide the larger number by the smaller one

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In how many ways can 6 students be chosen from 9 students if the order in which the

harkovskaia [24]

Answer:

60480

Step-by-step explanation:

This question has to do with permutations. A permutation is used to find the number of possible outcomes, assuming order matters. In this case, n=9 and r=6. This is written as $_{9}P_{6}$ which equals $\frac{9!}{(9-6)!}$ . Then, plug this into a calculator to get 60480.

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3 years ago

Solve for C = 2pie.

bonufazy [111]

I don’t think you can do that

5 0

3 years ago

Rectangle JKLM is shown.

ad-work [718]

Answer:

Step-by-step explanation:

We can use the Pythagorean theorem to solve this.

(13)^2 = (8)^2+x^2

169 = 64 + x^2

x^2 = 105

x is approximately 10.2, so B

6 0

3 years ago

SOMEONE PLEASEEEEW HELP ME OUTTT

garik1379 [7]

Answer:

By Double splitting method

=> 2x²-7x+5

=> 2x²-2x-5x+5

=>2x(x-1)-5(x-1)

x=5/2
x=1

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Hope it helps

-------☆ﾟ.･｡ﾟƒöᏝϗʆѻʁᶥąռ¹₃

4 0

3 years ago

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.

8 0

3 years ago