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DochEvi [55]
3 years ago
8

This table shows data collected by a runner.

Mathematics
1 answer:
kakasveta [241]3 years ago
8 0

Answer:

its the last anwser number 4

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How do I do this?<br>*Look at the directions in the photo*​
lora16 [44]

Answer:

Area\ of\ material\ required\ for\ the\ first\ box=384\ inches^2\\Area\ of\ material\ required\ for\ the\ second\ box=486\ inches^2\\Area\ of\ material\ required\ for\ the\ first\ box=600\ inches^2\\Total\ Area\ of\ material\ required=1470\ inches^2

Step-by-step explanation:

We\ are\ given:\\Diameter\ of\ the\ first\ volleyball=8\ inches \\Diameter\ of\ the\ second\ volleyball=9\ inches\\Diameter\ of\ the\ third\ volleyball= 10\ inches.\\Hence,\\We\ know\ that,\\If\ the\ side\ of\ the\ cube\ box\ is\ s, it's\ Total\ Surface\ Area\ =No.\ of\\ faces\ in\ a\ regular\ polyhedron\ *Area\ of\ each\ face\ of\ the\ polyhedron=6*s^2=6s^2\\Hence,\\Lets\ apply\ this\ equation\ in\ finding\ the\ area\ of\ material\ required\ for\ the\\ three\ cases.\\

As\ the\ volleyball\ should\ wholly\ fit\ into\ the\ box,\ the\ diameter\ of\ the\\ volleyballs\ would\ be\ the\ side\ of\ the\ cube\ box.\\Hence,\\For\ the\ first\ volleyball,\\Diameter\ of\ the\ first\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ first\ volleyball=8\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ first\ box=6s^2=6*8*8=384\ inches^2

For\ the\ second\ volleyball,\\Diameter\ of\ the\ second\ volleyball=9\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ second\ volleyball=9\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ second\ box=6s^2=6*9*9=486\ inches^2

For\ the\ third\ volleyball,\\Diameter\ of\ the\ third\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ third\ volleyball=10\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ third\ box=6s^2=6*10*10=600\ inches^2

Hence,\\If\ you\ are\ asked\ the\ Total\ Area\ to\ make\ all\ the\ boxes,\\ you\ just\ add\ them\ together.\\Hence,\\Total\ Area\ of\ Material\ required\ to\ make\ the\ three\ boxes=384+486+600=1470\ inches^2

7 0
3 years ago
giving the brainliest answer to who ever can help me out!!! AND has the correct answer need help asap!?!!
Afina-wow [57]

Answer:

caca water

pee water

Step-by-step explanation:

6 0
3 years ago
If f(x) = x + 4 and g(x)=x^2-1, what is m(g o f)(x)?
Andreas93 [3]

Until now, given a function  f(x), you would plug a number or another variable in for x. You could even get fancy and plug in an entire expression for x. For example, given  f(x) = 2x + 3, you could find f(y2 – 1) by plugging y2 – 1 in for x to get f(y2 – 1) = 2(y2 – 1) + 3 = 2y2 – 2 + 3 = 2y2 + 1.

In function composition, you're plugging entire functions in for the x. In other words, you're always getting "fancy". But let's start simple. Instead of dealing with functions as formulas, let's deal with functions as sets of (x, y) points:

Let f = {(–2, 3), (–1, 1), (0, 0), (1, –1), (2, –3)} and  

let g = {(–3, 1), (–1, –2), (0, 2), (2, 2), (3, 1)}.  

 

Find (i) f (1), (ii) g(–1), and (iii) (g o f )(1).

(i) This type of  exercise is meant to emphasize that the (x, y) points are really (x, f (x)) points. To find  f (1), I need to find the (x, y) point in the set of (x, f (x)) points that has a first coordinate of x = 1. Then f (1) is the y-value of that point. In this case, the point with x = 1 is (1, –1), so:

8 0
3 years ago
Read 2 more answers
Carl has a square table cloth with a total of 150 square feet. Which measurement is closest to the length of one side of the tab
Amanda [17]

Answer:

It will cost $190.75 to paint the whole fence.

Step-by-step explanation:

109 * 7 = 763 because Length x Width = Area

0.25 * 763 = 190.75

8 0
3 years ago
Determine the equations of the vertical and horizontal asymptotes, if any, for g(x)=x-2/x^2+4x+3
KatRina [158]

Answer:

<h2>B. x = -1, x = -3, y = 0</h2>

Step-by-step explanation:

g(x)=\dfrac{x-2}{x^2+4x+3}\\\\vertical\ asymptote:\\\\x^2+4x+3=0\\x^2+x+3x+3=0\\x(x+1)+3(x+1)=0\\(x+1)(x+3)=0\iff x+1=0\ \vee\ x+3=0\\\\\boxed{x=-1\ \vee\ x=-3}\\\\horizontal\ asymptote:\\\\\lim\limits_{x\to\pm\infty}\dfrac{x-2}{x^2+4x+3}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(\frac{1}{x}-\frac{2}{x^2}\right)}{x^2\left(1+\frac{4}{x}+\frac{3}{x^2}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{\frac{1}{x}-\frac{2}{x^2}}{1+\frac{4}{x}+\frac{3}{x^2}}=\dfrac{0}{1}=0\\\\\boxed{y=0}

4 0
3 years ago
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