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snow_lady [41]
3 years ago
7

Suppose the diameter of a circle is 4, What is its circumference?

Mathematics
1 answer:
weeeeeb [17]3 years ago
5 0
The circumference of a circle is pi(d) so if you multiply pi(4) = 3.141592654(4) you would get 12.5663706144
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There are 3 tennis balls in each can. The coach bought a total of 27 tennis balls. Write and solve the equation to find How many
natima [27]

Answer:He bought 9 cans.

Step-by-step explanation:

27divided by 3 is 9.

4 0
3 years ago
What is the ratio of a : b, knowing that a — b = 3 <br> a + b = 5
UNO [17]

Answer:

4:1

Step-by-step explanation:

a — b = 3

then a= 3+b

a + b = 5

3+b +b = 5

3+2b=5

2b=5-3

2b=2

b=1

a= 3+b

a=3+1=4

a:b= 4:1

3 0
3 years ago
you buy a new mountain bike for $200. tha value of the bike decreases by 25% each year. how much is your bike worth after 3 year
Brilliant_brown [7]

Answer:84.375

Step-by-step explanation:

i mean if you by a bike for 200 hundread and every year it goes down by 25% you would just subtract 25 % 3 times ;)

8 0
2 years ago
What is a - 5 - 6a + 3
Yakvenalex [24]

Answer:

-7-5a

Step-by-step

i simplified it if thats what it meant

4 0
3 years ago
Consider the inverse function. Which conclusions can be drawn about f(x) = x2 + 2? Select three options. f(x) has a limited rang
PolarNik [594]

Answer:

f(x) has a limited range

f(x) has a maximum at the point (0, 2)

f(x) has a y-intercept at the point (0, 2).

Step-by-step explanation:

Given the function;

f(x) = x^2+2

The domain is the value of the input variables for which the function will exist. According to the expression given, the function exists on all real values of x. The same goes with range which deals with the output values. It also exists on all real values from 2 and above.

Hence f(x) have a limited range (since values less than 2 are not included compare to domain that exists on all real values) and does not have a restricted domain.

For the x intercept, x intercept occur at y = 0

substitute y = 0 into the function and get y

if y = f(x)

y = x^2+2

0 = x^2 + 2

x^2 = -2

x = 2i

Hence  f(x) does not have an x-intercept of (2, 0)

For the y intercept, y intercept occur at x = 0

substitute x = 0 into the function and get y

if y = f(x)

y = x^2+2

y = 0^2 + 2

y = 2

Hence  f(x) has a y-intercept at point (0, 2)

f(x) is at maximum if d(fx))/dx = 0

d(fx))/dx  = 2x

since  d(fx))/dx  = 0

0 = 2x

x = 0

substitute x = 0 into the function

f(x) = x^2 + 2

y = 0^2+2

y = 2

Hence f(x) has a maximum at the point (0, 2)

5 0
2 years ago
Read 2 more answers
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