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Rudik [331]
3 years ago
6

Find the volume and thank for answer

Mathematics
2 answers:
Monica [59]3 years ago
5 0

Answer:

75.398

Step-by-step explanation:

shutvik [7]3 years ago
4 0

Answer: The volume of this cone is 75.4 m^3

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Estimate the distance you can travel in 3 hours 20 minutes if you drive on average 49 miles per hour. Round your answer to the n
AlexFokin [52]

Answer:

163 miles

Step-by-step explanation:

set up a ratio based on 49 miles in 1 hour (60 minutes):

let 'd' = distance in 3hrs, 20 min

3hrs, 20 min = 200 minutes

49/60 = d/200

cross-multiply:

60d = 9800

d = 163.3

7 0
2 years ago
Graph this on a coordinate plane f(x)=2x^2-1
rodikova [14]

Answer:

Step-by-step explanation:

7 0
2 years ago
How many multiples of 4 are there in {n; 37< n <1001}?
s2008m [1.1K]
Let's begin by listing the first few multiples of 4:  4, 8, 12, 16, 20, 24, 28, 32, 36, 38, 40, 44.  So, between 1 and 37 there are 9 such multiples:  {4, 8, 12, 16, 20, 24, 28, 32, 36}.  Note that 4 divided into 36 is 9.

Let's experiment by modifying the given problem a bit, for the purpose of discovering any pattern that may exist:

<span>How many multiples of 4 are there in {n; 37< n <101}?  We could list and then count them:  {40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100}; there are 16 such multiples in that particular interval.  Try subtracting 40 from 100; we get 60.  Dividing 60 by 4, we get 15, which is 1 less than 16.  So it seems that if we subtract 40 from 1000 and divide the result by 4, and then add 1, we get the number of multiples of 4 between 37 and 1001:

1000
   -40
-------
 960

Dividing this by 4, we get 240.  Adding 1, we get 241.

Finally, subtract 9 from 241:  We get 232.

There are 232 multiples of 4 between 37 and 1001.

Can you think of a more straightforward method of determining this number? </span>

5 0
3 years ago
I need help on this
Marina CMI [18]

I think it's answer choice a

3 0
3 years ago
Read 2 more answers
1+Sin/Cos + Cos/1+Sin = 2Sec
Bond [772]

Step-by-step explanation:

Consider the left-hand side:

\dfrac{1+\sin{\theta}}{\cos{\theta}} + \dfrac{\cos{\theta}}{1+\sin{\theta}}

\:\:\:\:= \dfrac{(1+\sin{\theta})^2 + \cos^2{\theta}}{\cos{\theta}(1+\sin{\theta})}

\:\:\:\:=\dfrac{1+2\sin{\theta}+\sin^2{\theta} + \cos^2{\theta}}{\cos{\theta}(1+\sin{\theta})}

\:\:\:\:=\dfrac{2+2\sin{\theta}}{\cos{\theta}(1+\sin{\theta})} =\dfrac{2(1+\sin{\theta})}{\cos{\theta}(1+\sin{\theta})}

\:\:\:\:= \dfrac{2}{\cos{\theta}} = 2\sec{\theta}

7 0
3 years ago
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