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bija089 [108]
3 years ago
13

What shape are possible cross section cylinder?

Mathematics
1 answer:
san4es73 [151]3 years ago
6 0

Answer:

Circle

Step-by-step explanation:

The cross-section of a cylinder is a circle.

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Solve -11 2/3 * (-4 1/5)
eduard

Answer:pts

Solve -11 2/3 * (-4 1/5)=49

Step-by-step explanation:

do whatever is in parentheses first

mark me brainest plz

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3 years ago
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Describe the interval(s) on which the function is continuous. (enter your answer using interval notation.)f(x) = xx + 6
Verdich [7]
F(x)=xx+6 = 1(x) =xx+7
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Which decimals in the list are repeating decimals? 1/3 3/5 4/15 5/6 4/1000 1 3/4
RSB [31]

Answer:

none

Step-by-step explanation:

1/3=0.33

3/5=0.6

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Not sure how to do #69, it's a calc 1 question
Gnoma [55]
Let A( t , f( t ) ) be the point(s) at which the graph of the function has a horizontal tangent => f ' ( t ) = 0.

But, f ' ( x ) = [ ( x^2 ) ' * ( x - 1 ) - ( x^2 ) * ( x - 1 )' ] / ( x - 1 )^2 =>
f ' ( x ) = [ 2x( x - 1 ) - ( x^2 ) * 1 ] / ( x - 1 )^2 => f ' ( x ) = ( x^2 - 2x ) / ( x - 1 )^2;

f ' ( t ) = 0 <=> t^2 - 2t = 0 <=> t * ( t - 2 ) = 0 <=> t = 0 or t = 2 => f ( 0 ) = 0; f ( 2 ) = 4 => A 1 ( 0 , 0 ) and A 2 ( 2 , 4 ).
3 0
3 years ago
If a person tosses a coin 23 times, how many ways can he get 11 heads
EastWind [94]

Tossing a coin is a binomial experiment.

Now lets say there are 'n' repeated trials to get heads. Each of the trials can result in either a head or a tail.

All of these trials are independent since the result of one trial does not affect the result of the next trial.

Now, for 'n' repeated trials the total number of successes is given by

_{r}^{n}\textrm{C}

where 'r' denotes the number of successful results.

In our case n=23 and r=11,

Substituting the values we get,

_{11}^{23}\textrm{C}=\frac{23!}{11!\times 12!}

\frac{23!}{11!\times 12!}=1352078

Therefore, there are 1352078 ways to get heads if a person tosses a coin 23 times.


3 0
4 years ago
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