Write a sequence of transformations that maps quadrilateral abcd onto quadrilateral a b c d

Mathematics

1 answer:

kakasveta [241]3 years ago

7 0

It can consist of translations (slides), reflections (flips), and rotations (turns).

I would say a reflection, then slide right, then slide down.

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Solve for p.<br> -p - 1 = 5 - 9p + 10p

Darina [25.2K]

Answer:

p= 1/3 or 0.33 repeating

Step-by-step explanation:

5-19p=-p-1

5=18p-1

6=18p

divide by 18

6/18 = 1/3 or 0.33 repeating

5 0

3 years ago

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quester [9]

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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

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

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