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astra-53 [7]
3 years ago
14

Find the circumstance of the circle with the radius of 6​

Mathematics
2 answers:
aleksandr82 [10.1K]3 years ago
5 0

Answer:

Step-by-step explanation:

c=2\pir

Inessa [10]3 years ago
3 0

Answer:

Circumference=37.6991 or 38

Step-by-step explanation:

c=2\pir

c=2\pi6

c=37.6991

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he measurements of two sides of a triangle are 4 cm and 20 cm. What could be a measurement of the third side of the triangle?
hodyreva [135]

Answer:

80

Step-by-step explanation:

every triangle adds up to 180 sort of, in this case, turn 4 into 40. 40+20=60 so add 80 to be 180!!!

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2 years ago
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3 years ago
What is the effect on the graph of the function f(x) = 2^x when f(x) is replaced with f(x − 3)?
mel-nik [20]

Graph of f(x-3) is compressed by a factor of  \frac{1}{8} horizontally of f(x).

<u>Step-by-step explanation:</u>

We have, the graph of f(x)= 2^{x} , on replacing f(x) by f(x-3) we get:

f(x-3)= 2^{x-3} = \frac{2^{x}}{2^{3}} = \frac{1}{8} 2^{x} = \frac{1}{8} f(x).Below shown are the images for graph of f(x) and f(x-3). Both are functions are exponential , and so having exponential graph but f(x-3) is compressed by a factor of  \frac{1}{8} horizontally . Domain and range of both functions are same i.e. F(x) & f(x-3) domain & range are same , just difference in graph : f(x-3) = \frac{1}{8} f(x).

4 0
3 years ago
30 mins = how many seconds???
horrorfan [7]
6x3=18 add two zeros and you get 1800 seconds:) hope this helped
8 0
3 years ago
Read 2 more answers
Solve by graphing: use any method<br><br> x + y = 6<br><br> x - y = 4
jekas [21]
X+y=6
x-y=4
using these equations we solve for either x or y in one and plug it into that variable in the other equation.
x=y+4 so y+4+y=6 so 2y=2 so y=1
now using y=1 put into first equation. x+1=6 so x=5 
so we know our graph goes through the coordinate point (5,1) then just make an x table and use some values to find the corresponding y coordinate and then do that for at least three points or more if needed to find the best fit for graph. Hope this helps walk you through it! Any questions please just ask! Thank you!
4 0
3 years ago
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