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Mrac [35]
3 years ago
12

David is filling a cylindrical cup with medication for his next patient. If he only fills the cup half way, what is the volume o

f medication in the cup if the radius is 2 inches and the height is 8 inches? 25.15 cubic inches 50.24 cubic inches 100.48 cubic inches 200.96 cubic inches
Mathematics
1 answer:
nignag [31]3 years ago
7 0

Answer:

Volume of medication in the cup = 50.24 cubic inches

Step-by-step explanation:

Volume of the cylindrical cup = πr^2h

Radius, r = 2 inches

Height, h = 8 inches

Volume of the cylindrical cup = πr^2h

= 3.14 × 2^2 × 8

= 3.14 × 4 × 8

= 100.48

Volume of the cylindrical cup = 100.48 cubic inches

If David only fills the cup half way,

then, the volume of medication in the cup = Total volume of the cylinder / 2

= 100.48 inches / 2

= 50.24 cubic inches

volume of medication in the cup = 50.24 cubic inches

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D
Gelneren [198K]

Answer:

General equation of line : y = mx+c   --1

Where m is the slope or unit rate

Table 1)

p    d

1      3

2     6

4     12

d = Number of dollars (i.e.y axis)

p = number of pound(i.e. x axis)

First find the slope

First calculate the slope of given points

m = \frac{y_2-y_1}{x_2-x_1}   ---A

(x_1,y_1)=(1,3)

(x_2,y_2)=(2,6)

Substitute values in A

m = \frac{6-3}{2-1}

m = 3

Thus the unit rate is 3 dollars per pound.

So, It matches the box 1 (Refer the attached figure)

Equation 1 : p=3d

\frac{p}{3}=d

Since p is the x coordinate and d is the y coordinate

On Comparing with 1

m = \frac{1}{3}

Thus the unit rate is \frac{1}{3} dollars per pound

So, It matches the box 2 (Refer the attached figure)

Equation 2 : \frac{1}{3}d=3p

d=9p

Since p is the x coordinate and d is the y coordinate

On Comparing with 1

m =9

Thus the unit rate is 9 dollars per pound

So, It matches the box 3 (Refer the attached figure)

Table 2)

p        d

1/9      1

1         9

2        18

d = Number of dollars (i.e.y axis)

p = number of pound(i.e. x axis)

(x_1,y_1)=(\frac{1}{9},1)

(x_2,y_2)=(1,9)

Substitute values in A

m = \frac{9-1}{1-\frac{1}{9}}

m = \frac{8}{frac{8}{9}}

m = 9

Thus the unit rate is 9 dollars per pound

So, It matches the box 3 (Refer the attached figure)

3 0
3 years ago
Read 2 more answers
Math6. Grade 6.
Iteru [2.4K]
2 (6 x+6)= ?
12 x+ 12=
12 x= -12
x= -1
So first you take what is outside the "( )" and you multiply it with what's in the "( )" so that gave us 12 because 2*6= 12. 
 
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3 years ago
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Given the geometric sequence where a1=3 and the common ratio is -1, what is the domain for n?
zimovet [89]

Answer:

  all integers where n ≥ 1

Step-by-step explanation:

n is a counting number, since it counts the terms of the sequence. It is in the set {1, 2, 3, ...}, any of the integers 1 or greater.

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3 years ago
What is the domain of the function continous or discrete
scoray [572]
The domain of the function is discrete 

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3 years ago
The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped
kakasveta [241]

Check the picture below.

\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2  = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

4 0
3 years ago
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