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Archy [21]
3 years ago
11

If a pizza pie had a diameter of 12 inches, what would be its circumference?

Mathematics
2 answers:
Ilya [14]3 years ago
6 0

Answer:

37.7 inches

Step-by-step explanation:

Formula of circumference:

2\pi r

r = radius

In the current case,

radius is 6 inches

Hence,

circumference=

2*\pi *6\\= 12\pi\\= 37.6991118431

≅ 37.7 inches

<em>Feel free to mark it as brainliest</em>

fenix001 [56]3 years ago
3 0
Circumference = π x d
= π x 12
= 37.699
Rounded off is 37.70
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Elza [17]

Answer:

linear

Step-by-step explanation:

if you have access to a graphing calculator then enter the equation and you can see it forms a line

if you have graph paper, create some coordinates, plot and connect them

example of some ordered pairs could be:

(0, 7)

(2, 6)

(-2, 8)

3 0
2 years ago
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45 POINTS PLEASE HELP!
andrew-mc [135]

Hello,

Question:

Jack popped some popcorn, but 3 of the 150 kernels did not pop.

Determine which popcorn snacks below has the same ratio of unpopped to total kernels as Jack's popcorn.

We Know:

3 out of 150 did not pop

Solution:

3/150 = 3 to 150

Find ratios similar to this one,

4/200

2/100

1/50

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4 unpopped out of 200 kernels

1 unpopped out of 50 kernels

8 0
3 years ago
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mrs_skeptik [129]
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4 0
3 years ago
the school fun fair made 1,768 on games and 978 on food sales. how much money did the fun fair make on games and food.
Roman55 [17]
The school made 2746 dollars. 
4 0
3 years ago
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If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf
jonny [76]
With the curve C parameterized by

C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k

with 0\le t\le\dfrac\pi2, and given the vector field

\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k

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\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt

where

\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt

The integral is then

\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt
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=269
6 0
3 years ago
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