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Sphinxa [80]
3 years ago
13

Write an algebraic rule for the rotation for s (-1,3) s' (-3,-1)

Mathematics
1 answer:
USPshnik [31]3 years ago
6 0
That's the symmetric property
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Solve questions 15,16,17, and 18. Will give Brainliest!! Please!!! Show minimal work!
SVEN [57.7K]

15. The terminal side of \frac{3\pi}{2} intercepted the unit circle at:

(0,-1).

This implies that:

(\cos \frac{3\pi}{2}, \sin \frac{3\pi}{2}=(0,-1)

This implies that:

\cos \frac{3\pi}{2}=0,\:and\: \sin \frac{3\pi}{2}=-1

16) We have that: \cot \theta =-\frac{6}{7}.

We illustrate this on the right triangle and apply the Pythagoras Theorem as follows:

h^2=6^2+7^2

h^2=36+49

h^2=85

h=\sqrt{85}

Using the mnemonics SOH-CAH-TOAH, we have:

\sin \theta=-\frac{Opp}{Hyp}=-\frac{7}{\sqrt{85} }=-\frac{7\sqrt{85} }{85}

\cos \theta=\frac{Adj}{Hyp}=\frac{6}{\sqrt{85} }=\frac{6\sqrt{85} }{85}

\tan \theta=-\frac{Opp}{Adj}=-\frac{7}{6}}

\csc \theta=-\frac{Hyp}{Opp}=-\frac{\sqrt{85} }{7}

\sec \theta=\frac{Hyp}{Adj}=\frac{\sqrt{85} }{6}

17. We want to verify that: sin^4x-sin^2x=cos^4x-cos^2x

Verifying from the LHS

\sin^4x-sin^2x=(sin^2x)^2-sin^2x

Recall that from the Pythagorean identity:\sin^2x=1-\cos^2x

\sin^4x-sin^2x=(1-cos^2x)^2-(1-\cos^2x)

\sin^4x-sin^2x=1-2cos^2x+\cos^4x-1+\cos^2x

\sin^4x-sin^2x=\cos^4x-\cos^2x

18.  We have:

\tan^2x\sec^2x+2\sec^2x-\tan^2x=2

\tan^2x\sec^2x+2\sec^2x-\tan^2x-2=0

\sec^2x(\tan^2x+2)-1(\tan^2x+2)=0

(\sec^2x-1)(\tan^2x+2)=0

When (\sec^2x-1)=0, we have

x=0,\pi

When \tan^2x+2=0, \tan^2x=-2

We can see this is not defined for all real values of x.

6 0
3 years ago
Question 4 (4 points
aalyn [17]

Answer:

It is True 200% legit!

Step-by-step explanation:

4 0
3 years ago
1/6x -5/9xy^2<br> Find the least common denominator for the two expressions.
eimsori [14]

Answer:

Step-by-step explanation:

6x=2×3×x

9xy²=3×3×x×y²

common denominator=2×3×3×x×y²=18xy²

4 0
4 years ago
In the following diagram,
olga2289 [7]

Answer:

  • x + 75 = 180, as these two angles form a straight angle.
  • x = 180° - 75° = 105°

<u>y and x are supplementary angles as p is transversal of l and m:</u>

  • y + x = 180°
  • y = 180° - x = 180°  - 105° = 75°

<u>z is same as 86° as vertical angles:</u>

  • z = 86°

<u>w and z are supplementary angles as g is transversal of l and m:</u>

  • w + z = 180°
  • w = 180° - z = 180° - 86° = 94°
6 0
3 years ago
Juno calculated the area of a square to be 4/9 square yard. Which shows the side length of the square?
Stella [2.4K]
\bf \textit{area of a square}\\\\&#10;A=s^2~~&#10;\begin{cases}&#10;s=side's~length\\&#10;--------\\&#10;A=\frac{4}{9}&#10;\end{cases}\implies \cfrac{4}{9}=s^2&#10;\\\\\\&#10;\sqrt{\cfrac{4}{9}}=s\implies \cfrac{\sqrt{4}}{\sqrt{9}}=s\implies \cfrac{2}{3}=s
3 0
3 years ago
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