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Leno4ka [110]
3 years ago
6

What is the circumference of the circle below? (Round your answer to the nearest tenth.)

Mathematics
1 answer:
yuradex [85]3 years ago
3 0

Answer:

D

Step-by-step explanation:

Circumference of circle = 2πr

r = 9cm

Circumference of circle = 2×9×3.142

Circumference of circle = 56.5cm

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3 * x - 4 equals 2 * -2x + 1
Crank

Answer:

x=2

Step-by-step explanation:

3(x-4) = 2(-2x+1) [Distribute]

-> 3x - 12 = -4x + 2 [Simplify]

-> 7x = 14 [Divide by 7 on both sides]

-> x = 2

6 0
3 years ago
Based only on the given information, it is guaranteed that
Kaylis [27]

Answer:

your answer is true

Step-by-step explanation:

both angles at the start of the triangle are the sames so the distance between the center line to the outsides are the same

5 0
3 years ago
Find the equation of the line passing through 2,11
Cerrena [4.2K]

Answer:

Step-by-step explanation:

There is an infinite number of lines that pass through the point (2, 11). Therefore, there is an infinite number of equations. To define a single line, you must have at least two points. One point is not enough.

3 0
3 years ago
The perimeter of a rectangle is 52 feet. describe the possible lengths of a side if the area of the rectangle is not to exceed 1
abruzzese [7]

25* 1

24* 2

23* 3

22* 4

21* 5

20* 6

and finally 19* 7

so in all, there are 7 solutions.

5 0
3 years ago
What is the relative maximum and minimum of the function?
garri49 [273]
Take the derivitive
f'(x)=6x^2+2x-11

find where f'(x)=0
f'(x)=0 when x=-1.53089 or x=1.19756

we use a sign chart
test values to see where the signs are
(see attachment)
f'(-2)=(+)
f'(0)=(-)
f'(2)=(+)
max happens when sign changes from (+) to (-)
min happens when sign changes from (-) to (+)

according to the chart, max is at -1.53089 and min is at 1.19756

now evaluate the original function for x=-1.53089 and x=1.19756
f(-1.53089)=12.0078
f(1.19756)=-8.30405

max at (-1.53089,12) and min at (1.19756, -8.30405)
I may have rounded off differently, but


answer is 2nd option

5 0
3 years ago
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