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ki77a [65]
3 years ago
15

The diameter of a circle is 3 meters. What is the circumference?

Mathematics
1 answer:
Gennadij [26K]3 years ago
3 0

Answer:

C≈9.42m

Using any of this formulas

C=2πr

C=2πrd=2r

Solution:

C=πd=π·3≈9.42478m

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Identify the terms and like terms in the expression.
AleksandrR [38]

Answer:

terms: 2n, -n, -4, 7n

like terms: 2n, -n, 7n

Step-by-step explanation:

4 0
2 years ago
What is 1 1/3 divided by 3/5
enyata [817]

Answer:

55/9

Step-by-step explanation:

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3 years ago
Read 2 more answers
Help needed desperately
Aneli [31]

Answer:

80° , 100°, 80°

Step-by-step explanation:

∠ 1 and ∠ 2 form a straight line and sum to 180° , then

2x + 40 + 2y + 40 = 180

2x + 2y + 80 = 180 ( subtract 80 from both sides )

2x + 2y = 100 → (1)

∠ 1 and ∠ 3 are vertical angles and are congruent , then

2x + 40 = x + 2y ( subtract x from both sides )

x + 40 = 2y ( subtract 40 from both sides )

x = 2y - 40 → (2)

Substitute x = 2y - 40 into (1)

2(2y - 40) + 2y = 100

4y - 80 + 2y = 100

6y - 80 = 100 ( add 80 to both sides )

6y = 180 ( divide both sides by 6 )

y = 30

Substitute y = 30 into (2)

x = 2(30) - 40 = 60 - 40 = 20

Thus x = 20 and y = 30

Then

∠ 1 = 2x + 40 = 2(20) + 40 = 40 + 40 = 80°

∠ 2 = 2y + 40 = 2(30) + 40 = 60 + 40 = 100°

∠3 = x + 2y = 20 + 2(30) = 20 + 60 = 80°

8 0
3 years ago
HELP!!! please this is my last question for my paper
s2008m [1.1K]

Answer:

\sqrt{2}

Step-by-step explanation:

We can use pytho theorem to find X because its a right triangle, so:

a^2 + b^2 = c^2

plug our numbers

1^2 + 1^2 = x^2

simplify

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4 0
2 years ago
Answer the following question about the function whose derivative is given below
Komok [63]

Answer:

a) The critical points are x = 3 and x = -6.

b) f is decreasing in the interval (-\infty, -6)

f is increasing in the intervals (-6,3) and (3,\infty).

c) Local minima: x = -6

Local maxima: No local maxima

Step-by-step explanation:

(a) what are the critical points of f?

The critical points of f are those in which f^{\prime}(x) = 0. So

f^{\prime}(x) = 0

(x-3)^{2}(x+6) = 0

So, the critical points are x = 3 and x = -6.

(b) on what intervals is f increasing or decreasing? (if there is no interval put no interval)

For any interval, if f^{\prime} is positive, f is increasing in the interval. If it is negative, f is decreasing in the interval.

Our critical points are x = 3 and x = -6. So we have those following intervals:

(-\infty, -6), (-6,3), (3, \infty)

We select a point x in each interval, and calculate f^{\prime}(x).

So

-------------------------

(-\infty, -6)

f^{\prime}(-7) = (-7-3)^{2}(-7+6) = (100)(-1) = -100

f is decreasing in the interval (-\infty, -6)

---------------------------

(-6,3)

f^{\prime}(2) = (2-3)^{2}(2+6) = (1)(8) = 8

f is increasing in the interval (-6,3).

------------------------------

(3, \infty)

f^{\prime}(4) = (4-3)^{2}(4+6) = (1)(10) = 10

f is increasing in the interval (3,\infty).

(c) At what points, if any, does f assume local maximum and minima values. ( if there is no local maxima put mo local maxima) if there is no local minima put no local minima

At a critical point x, if the function goes from decreasing to increasing, it is a local minima. And if the function goes from increasing to decreasing, it is a local maxima.

So, for each critical point is this problem:

At x = -6, f goes from decreasing to increasing.

So x = -6, f assume a local minima value

At x = 3, f goes from increasing to increasing. So, there it is not a local maxima nor a local minima. So, there is no local maxima for this function.

4 0
3 years ago
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