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Burka [1]
3 years ago
13

Two pizzas with 8 inch and 16 inch diameters are each cut into 6 equal pieces. How does the area of each piece of the smaller pi

zza compare to the area of each piece of the larger pizza?
Mathematics
1 answer:
trasher [3.6K]3 years ago
6 0

Answer:

Step-by-step explanation:

The easiest way to think about increasing area is by taking a square, if you double the dimensions, let's say they were 1 by 1 and then was made into a 2x2 square, you can fit four of the original cube (1x1) into the 2x2. So when the dimensions are doubled the area is quadrupled.

Even easier shortcut:

Take the cube idea, when the dimensions are messed with for anything, (works for volume too) always think of the smaller piece as 1x1 or 1x1x1. Then if the dimensions are double make it 2x2 or 2x2x2. If the dimensions are tripled then do 3x3 or 3x3x3, etc... 1x1=1x1x1, they are both equal to 1. So think of that as the numerator. the 2x2 in this case is equal to 4 so the smaller piece is 1/4 of the bigger one. Shown with volume if you double the dimensions of a cube that was 1x1x1 into 2x2x2. 1x1x1 still equals 1 so that's still the numerator and 2x2x2 is equal to 8 so the 1x1x1 cube is 1/8 of the 2x2x2, in other words you can fit eight 1x1x1 cubes into the 2x2x2 cube.

Hope this helps with area and volume.

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Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
7 0
3 years ago
Find the distance between the two points rounding to the nearest tenth (if necessary).
Virty [35]

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{3}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{9})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{(1-3)^2+(9-6)^2}\implies d=\sqrt{(-2)^2+3^2} \\\\\\ d=\sqrt{13}\implies d\approx 3.6056\implies \stackrel{\textit{rounded up}}{d=3.6}

4 0
4 years ago
2) 83,97,85,84,96, 80,80,87,91<br> Mean median and mode
marusya05 [52]

\text {Hi! Let's Solve this Problem!}

\text {Before you find the Mean, Medina and Mode you must put the numbers in order.}

\text {Before: 83, 97, 85, 84, 96, 80, 80, 87, 91}\\\text {After: 80, 80, 83, 84, 85, 87, 91, 96, 97}

\underline {\text {Mean}}

\text {To Find the Mean you must Add all the Numbers then Divide.}

\text {Add:} \text 80+80+83+84+85+87+91+96+97=\fbox {783}}

\text {Divide:} \text {783/9=}

\text {The Mean Is:}

\fbox {87}

\underline {\text Median}}

\text {To find the Median you find the number that's in the middle of the set.}

\text {Put your Left Pointer Finger on 83. Put your Right Pointer Finger on 91.}

\text {Move your Left Finger to the Right. Move your Right Finger to the Left.}

\text {Once your Left Finger makes it to 84 and your Right Finger makes it to 87} \text {it shows that 85 is left in the middle.}

\text {The Median Is:}

\fbox {85}

\underline {\text {Mode}}

\text {Finding the Mode means finding the number} \text { that shows the most in the number set given.}

\text {Since 80 is the number that is showing more than any other number} \text {this tells us that 80 is our Mode.}

\text {The Mode Is:}

\fbox {80}

\text {Best of Luck!}

4 0
4 years ago
Correct/best answer will get brainliest!
vodka [1.7K]
The answer would be b.100 because its clearly not a right triangle
6 0
4 years ago
Solve the linear equation 5(8x+2) -64=2(8x9) Find X
raketka [301]
Bla bla bla your answer is x=4.95

3 0
3 years ago
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