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Oksana_A [137]
4 years ago
9

geometry question(s) on triangle similarity, please help, i’d really appreciate it! image is included.

Mathematics
1 answer:
Sergio [31]4 years ago
5 0
1. D. The triangles have two given sides with an included angle, so the postulate would be SAS.

2. A. The triangles have three given sides, so the postulate would be SSS.
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Answer: 5,000,000+600,000+70,000+8,000+200+9

Step-by-step explanation:

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Which expression is equivalent
VLD [36.1K]

Answer:

B)

Step-by-step explanation:

You must divide these fractions with square roots:

18x^{2}\sqrt{14x^8} / 6\sqrt{7x^4} = 3x^{2} \sqrt{2x^4} = 3x^{2}\sqrt{2}x^{2} = 3x^4\sqrt{2}

8 0
3 years ago
Jack is investing $3,000 into an account
marin [14]

Answer:

Step-by-step explanation:

I think this is the equation for annual compunt interest

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FV - PV = Interest

                           FV                   -      PV

Interest  = [3000(1 + 0.03)^30]   -   3000

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5 0
3 years ago
For a propular Broadway musical the theater box office sold 356 tickets at $80 apiece,275 tickets for $60 apiece and 369 tickes
Alenkasestr [34]
The first step to solving this equation would be to how much they made for every apiece. To do this we must multiply the cost of the $80 apiece by how much they sold, as well multiplying the $45 apiece by how many they sold. And also the $60 apiece by the how many they sold.

$80 apiece × 356 tickets = ?

$60 apiece × 275 tickets = ?

$45 apiece × 369 tickets = ?

Now we must solve our equations.

$80 apiece × 356 tickets = $28,480

$60 apiece × 275 tickets = $16,500

$45 apiece × 369 tickets = $16,605

The next step would be to add up all the total costs of each apiece.

$28,480 + $16,500 + $16,605 = $61,585

So our answer is...

The box office took in a total of $61,585
6 0
4 years ago
If sin theta = 2/5 and theta is in quadrant I, determine the following.
anzhelika [568]

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


recall that cosine is positive on the I Quadrant, so though we get a ± valid roots, only the positive one applies.


\bf cos(\theta)=\cfrac{\stackrel{adjacent}{\sqrt{21}}}{\stackrel{hypotenuse}{5}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{2}}{\stackrel{adjacent}{\sqrt{21}}} \qquad \qquad cot(\theta)=\cfrac{\stackrel{adjacent}{\sqrt{21}}}{\stackrel{opposite}{2}} \\\\\\ csc(\theta)=\cfrac{\stackrel{hypotenuse}{5}}{\stackrel{opposite}{2}} \qquad \qquad sec(\theta)=\cfrac{\stackrel{hypotenuse}{5}}{\stackrel{adjacent}{\sqrt{21}}}


now, for tangent and secant, let's rationalize the denominator.


\bf tan(\theta)\implies \cfrac{\stackrel{opposite}{2}}{\stackrel{adjacent}{\sqrt{21}}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{21} \\\\\\ sec(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\stackrel{adjacent}{\sqrt{21}}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{5\sqrt{21}}{21}

5 0
3 years ago
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