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3 years ago

To be an angle in quadrant II such that cos O= -5/8 Find the exact values of csc O and cot O.

1 answer:

5 0

let's recall that the cosine/adjacent is negative on the II Quadrant, whilst the sine/opposite is positive on that same quadrant, also let's recall that the hypotenuse is never negative, since it's just a radius distance.

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{-5}}{\stackrel{hypotenuse}{8}}\qquad \impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{8^2-(-5)^2}=b\implies \pm\sqrt{64-25}=b\implies \pm\sqrt{39}=b\implies \stackrel{II~Quadrant}{+\sqrt{39}=b} \\\\[-0.35em] ~\dotfill$

$\bf csc(\theta )\implies \cfrac{\stackrel{hypotenuse}{8}}{\stackrel{opposite}{\sqrt{39}}}\implies \cfrac{8}{\sqrt{39}}\cdot \cfrac{\sqrt{39}}{\sqrt{39}}\implies \cfrac{8\sqrt{39}}{39} \\\\\\ cot(\theta )\implies \cfrac{\stackrel{adjacent}{-5}}{\stackrel{opposite}{\sqrt{39}}}\implies \cfrac{-5}{\sqrt{39}}\cdot \cfrac{\sqrt{39}}{\sqrt{39}}\implies \cfrac{-5\sqrt{39}}{39}$

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3 years ago

Without plotting points, let M=(-2,-1), N=(3,1), M'= (0,2), and N'=(5, 4). Without using the distanceformula, show that segments

kramer

Given:

M=(x1, y1)=(-2,-1),

N=(x2, y2)=(3,1),

M'=(x3, y3)= (0,2),

N'=(x4, y4)=(5, 4).

We can prove MN and M'N' have the same length by proving that the points form the vertices of a parallelogram.

For a parallelogram, opposite sides are equal

If we prove that the quadrilateral MNN'M' forms a parallellogram, then MN and M'N' will be the oppposite sides. So, we can prove that MN=M'N'.

To prove MNN'M' is a parallelogram, we have to first prove that two pairs of opposite sides are parallel,

Slope of MN= Slope of M'N'.

Slope of MM'=NN'.

$\begin{gathered} \text{Slope of MN=}\frac{y2-y1}{x2-x1} \\ =\frac{1-(-1)}{3-(-2)} \\ =\frac{2}{5} \\ \text{Slope of M'N'=}\frac{y4-y3}{x4-x3} \\ =\frac{4-2}{5-0} \\ =\frac{2}{5} \end{gathered}$

Hence, slope of MN=Slope of M'N' and therefore, MN parallel to M'N'

$\begin{gathered} \text{Slope of MM'=}\frac{y3-y1}{x3-x1} \\ =\frac{4-(-1)}{5-(-2)} \\ =\frac{3}{2} \\ \text{Slope of NN'=}\frac{y4-y2}{x4-x2} \\ =\frac{4-1}{5-3} \\ =\frac{3}{2} \end{gathered}$

Hence, slope of MM'=Slope of NN' nd therefore, MM' parallel to NN'.

Since both pairs of opposite sides of MNN'M' are parallel, MM'N'N is a parallelogram.

Since the opposite sides are of equal length in a parallelogram, it is proved that segments MN and M'N' have the same length.

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1 year ago

Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the

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Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees was $x+\frac{1}{4} x=\frac{5}{4} x$ , and for the next three years we have that

Start End

Second year $\frac{5}{4}x$ -------------- $\frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x$

Third year $(\frac{5}{4} )^{2}x$ ------------- $(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x$

Fourth year $(\frac{5}{4})^{3}x$ -------------- $(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.$

So the formula to calculate the number of trees in the fourth year is

$(\frac{5}{4} )^{4} x,$ we know that all of the trees thrived and there were 6250 at the end of 4 year period, then

$6250=(\frac{5}{4} )^{4}x$ ⇒ $x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.$

Therefore the number of trees at the begging of the 4-year period was 2560.

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3 years ago