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3 years ago

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:

8 0

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

1+1 = x^2

2=x^2

answer = root2

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 $f^{\prime}(x)$ .

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

$(-\infty, -6)$

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

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