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diamong [38]
3 years ago
13

K is no greater than 9? Write>,<.

Mathematics
1 answer:
alekssr [168]3 years ago
4 0
Greater than symbol is ">"
But since he mentioned "NO greater than" then it's the opposite of greater than, which means "K" is less than "9"

K<9
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Fatima writes down three square numbers.
nekit [7.7K]

Answer:

10

Step-by-step explanation:

Fatima wrote down 3 square numbers

since we don't know the three square we will represent it by any variable (X)

3x=30

so you divide both sides by 3

so we have 3x/3=30/3=10

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3 years ago
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

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

h^2=6^2+7^2

h^2=36+49

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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}

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\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

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18.  We have:

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

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\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

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6 0
3 years ago
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Gre4nikov [31]

Answer:P=18cm/A=18cm

Step-by-step explanation:Trust me it's right

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3 years ago
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lyudmila [28]

Answer:

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Step-by-step explanation:

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daser333 [38]

Answer:

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The graph is widened.

The graph is shifted left 3 units.

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