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Schach [20]
3 years ago
6

PLEASE HELP ME :(( i tried 12 as the answer and it’s wrong

Mathematics
2 answers:
AnnZ [28]3 years ago
6 0

I think its 1/6

I'm not sure tho

u take the bottom and subtract it by the top then multiply

andrey2020 [161]3 years ago
5 0

Answer:

I think its 1/6

You will end up taking the bottom number and subtract it by the top number then multiply to get your final

Step-by-step explanation:

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Someone please help me!!
mote1985 [20]

Answer:B and E

Step-by-step explanation:

5 0
2 years ago
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What is the approximate percent decrease when a fish count goes down from 850 to 500? The percent decrease is approximately 40%.
Korolek [52]

Answer:

<h2><em>The percent decrease is approximately 40%. </em></h2>

Step-by-step explanation:

Step one:

given data

initial count of fish= 850

final count of fish= 500

Required

% decrease

Step two:

% decrease= final-initial/initial*100

substitute

% decrease= 850-500/850*100

% decrease=350/850*100

% decrease= 0.41*100

% decrease= 40.1%

<em>The percent decrease is approximately 40%. </em>

7 0
3 years ago
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Find x<br> a) 21 √2<br> b)7<br> c)21 √2/2<br> d)21 √3/2
fgiga [73]

Answer:

C

Step-by-step explanation:

Use the sine ratio in the left, right triangle to find the common side to both triangles and the exact values

sin60° = \frac{\sqrt{3} }{2} , cos45° = \frac{1}{\sqrt{2} } , thus

sin60° = \frac{opposite}{hypotenuse} = \frac{opp}{7\sqrt{3} } = \frac{\sqrt{3} }{2} ( cross- multiply )

2 × opp = 21 ( divide both sides by 2 )

opp = \frac{21}{2}

Now consider the right triangle on the right, using the cosine ratio

cos45° = \frac{adjacent}{hypotenuse} = \frac{\frac{21}{2} }{x} = \frac{1}{\sqrt{2} } ( cross- multiply )

x = \frac{21}{2} × \sqrt{2} = \frac{21\sqrt{2} }{2} → C

5 0
2 years ago
Lucy’s house is located at the point shown on the coordinate grid. Ainsley’s house is located 2 units right and 3 units down fro
dybincka [34]

Answer: 3,3

Step-by-step explanation:

5 0
2 years ago
RIP OpenStudy ;(
klemol [59]
First note that \frac{2^n+1}{2^{n+1}} =  \frac{2^n}{2^{n+1}} + \frac{1}{2^{n+1}} = \frac{1}{2} + \frac{1}{2^{n+1}}

If you take limit, then you have \lim_{n \to \infty}( \frac{1}{2} + \frac{1}{2^{n+1}})= \lim_{n \to \infty}( \frac{1}{2}) +\lim_{n \to \infty}(\frac{1}{2^{n+1}})=\frac{1}{2} +0= \frac{1}{2}



3 0
2 years ago
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