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3 years ago

Which formula could be used to determine the circuference of a circle

Mathematics

1 answer:

tia_tia [17]3 years ago

7 0

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ The formula for this would be:

2пr

In words, this would be two x pi x radius

[note: the radius is half the diameter]

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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A toy baseball bat comes with 3 plastic balls. The balls are in a box that is a rectangular solid. The box is just big enough to

vodomira [7]

Answer:

31.4 in³

Step-by-step explanation:

The box is just big enough to hold the 3 balls, so it must have a length 6 times the radius of each ball, a width 2 times the radius, and a height 2 times the radius.

The volume of the box is:

V = (6r)(2r)(2r)

V = 24r³

The volume of the 3 balls is:

V = 3 (4/3 π r³)

V = 4πr³

So the volume of the air is:

V = 24r³ − 4πr³

V = (24 − 4π) r³

Since r = 1.4 inches:

V = (24 − 4π) (1.4 in)³

V ≈ 31.4 in³

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3 years ago

1.1.31

miv72 [106K]

Answer:yes

Step-by-step explanation:well its just yes so your welcome

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4 years ago

MARK AS BRAINLEST

Art [367]

Answer:

Step-by-step explanation:

Good evening ,

I guess it’s the C expression,

Because, C. 3n^2+n-2 so the coefficient of the term n is 1 and the coefficient of the term n^2 is 3

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4 years ago

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Question 2b only! Evaluate using the definition of the definite integral(that means using the limit of a Riemann sum

lara [203]

Answer:

Hello,

Step-by-step explanation:

We divide the interval [a b] in n equal parts.

$\Delta x=\dfrac{b-a}{n} \\\\x_i=a+\Delta x *i \ for\ i=1\ to\ n\\\\y_i=x_i^2=(a+\Delta x *i)^2=a^2+(\Delta x *i)^2+2*a*\Delta x *i\\\\\\Area\ of\ i^{th} \ rectangle=R(x_i)=\Delta x * y_i\\$

$\displaystyle \sum_{i=1}^{n} R(x_i)=\dfrac{b-a}{n}*\sum_{i=1}^{n}\ (a^2 +(\dfrac{b-a}{n})^2*i^2+2*a*\dfrac{b-a}{n}*i)\\$

$=(b-a)^2*a^2+(\dfrac{b-a}{n})^3*\dfrac{n(n+1)(2n+1)}{6} +2*a*(\dfrac{b-a}{n})^2*\dfrac{n (n+1)} {2} \\\\\displaystyle \int\limits^a_b {x^2} \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} R(x_i)\\\\=(b-a)*a^2+\dfrac{(b-a)^3 }{3} +a(b-a)^2\\\\=a^2b-a^3+\dfrac{1}{3} (b^3-3ab^2+3a^2b-a^3)+a^3+ab^2-2a^2b\\\\=\dfrac{b^3}{3}-ab^2+ab^2+a^2b+a^2b-2a^2b-\dfrac{a^3}{3} \\\\\\\boxed{\int\limits^a_b {x^2} \, dx =\dfrac{b^3}{3} -\dfrac{a^3}{3}}\\$

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3 years ago

Let c = 11. What is the value of 6c?

Aleonysh [2.5K]

6 times 11 = 66 because it is 6c. That is your answer. I hope this helps. Could I possibly get brainliest?

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4 years ago

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