Because they are all divided by 8 so you can put 2, 5, and 1 in the order so you will have:
5/8 > 2/8 > 1/8
so 5/8 is the greatest :)))
i hope this is helpful
have a nice day

7 0

3 years ago

Read 2 more answers

Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.

vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

$cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}$

$\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}$

4 0

2 years ago

Intr un album numarul de fotografii s a micsorat cu 50%. Cu cat trebuie sa se mareasca pentru a reveni la numarul initial? Raspu

hodyreva [135]

Răspuns:

100%

Explicație pas cu pas:

Permiteți numărul inițial de fotografii = x

x - 50% din x = x - 0,5x = 0,5x

0,5x + câte% din 0,5x = x

x - 0,5x = y% de 0,5x

0,5x / 0,5x = 1

Prin urmare,

100% = 1

Prin urmare, x trebuie mărit cu 100% pentru a reveni la numărul original.

3 0

3 years ago