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Alekssandra [29.7K]
4 years ago
12

Which of the following formulas would find the surface area of a right cylinder where h is the height, r is the radius, LA is th

e lateral area, and BA is the base area?
A. 2pi^2+pi r h
B. BA+La
C. BA+ 2 pi r ^2
D. Pi r^2+ pi r h
E. LA+ (pi r^ 2)
Mathematics
1 answer:
Sedbober [7]4 years ago
8 0
I'd say it is A but I think you have made a tpyo
It should be 2*PI*radius^2 + 2*PI*radius*height


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Answer:

it is absolutely true:)

hope this helps

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WILLL GIVE BRAINLIEST + 50 POINTS FOR HELP!!!!<br><br><br>PLZZZZZZ IM TIMED
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A special deck of cards has ten cards. Four are green, three are blue, and three are red. When a card is picked, its color of it
nata0808 [166]

Answer:

P(A) = 3/20

Step-by-step explanation:

P(A)=P(blue)P(head)=(3/10)(1/2)=3/20

as there are 10 cards in total, out of which 3 are blue so the probability to get the blue card is, P(blue) = 3/10. and the probability of getting a head when a coin is tossed is P(head) = 1/2.

So in total

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3 years ago
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NARA [144]

Ans(a):

Given function is f(x)=\frac{3x-1}{x+4}

we know that any rational function is not defined when denominator is 0 so that means denominator x+4 can't be 0

so let's solve

x+4≠0 for x

x≠0-4

x≠-4

Hence at x=4, function can't have solution.


Ans(b):

We know that vertical shift occurs when we add something on the right side of function so vertical shift by 4 units means add 4 to f(x)

so we get:

g(x)=f(x)+4

g(x)=\frac{3x-1}{x+4}+4

We may simplify this equation but that is not compulsory.

Comparision:  

Graph of g(x) will be just 4 unit upward than graph of f(x).


Ans(c):

To find value of x when g(x)=8, just plug g(x)=8 in previous equation

8=\frac{3x-1}{x+4}+4


8-4=\frac{3x-1}{x+4}


4=\frac{3x-1}{x+4}


4(x+4)=(3x-1)


4x+16=3x-1


4x-3x=-1-16

x=-17

Hence final answer is x=-17

8 0
4 years ago
The volume of a box is 4 in. wide, 2 in high and 5 in. long is
sladkih [1.3K]

Answer:

40 in^3

Step-by-step explanation:

Volume of a box is given by

V = l*w*h

V = 4*2*5

   = 40 in^3

8 0
4 years ago
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