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Vikentia [17]
3 years ago
6

WILL GIVE BRAINLEIST! :Given a traffic cone has a base diameter of 12 inches and a height of 28 inches, which is the closest vol

ume of the cone?
 A- 2,463.01 in33

B- 3,166.73 in33

C- 4,222.3 in33

D-  1,055.58 in3

​
Mathematics
1 answer:
lesya [120]3 years ago
3 0

Answer:

1,055.58 in^3

Step-by-step explanation:

The volume of a cone is given by

V = 1/3 pi r^2 h

We need the radius

r = d/2 = 12/2 =6

V = 1/3 (pi) (6)^2 *28

Using the pi button on the calculator

   =1055.575132

Rounding to 2 decimal places

 1,055.58 in^3

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2 years ago
Use the equation a = IaIâ
german

Answer:

a) \:\:=\sqrt{14}\cdot \frac{\:\:}{\sqrt{14} }

b)\:\:=\sqrt{29} \cdot \frac{\:\:}{\sqrt{29} }

c) \:\:=7\cdot \frac{\:\:}{7}

Step-by-step explanation:

a) Let <u>a</u>=<2,1,-3>

The magnitude of <u>a</u> is |a|=\sqrt{2^2+1^2+(-3)^2}

|a|=\sqrt{4+1+9}=\sqrt{14}

The unit vector in the direction of a is

\hat{a}=\frac{\:\:}{\sqrt{14} }

Using the relation a=|a|\hat{a}, we have

\:\:=\sqrt{14}\cdot \frac{\:\:}{\sqrt{14} }

b) Let a=2i - 3j + 4k

|a|=\sqrt{2^2+(-3)^2+4^2}

|a|=\sqrt{4+9+16}=\sqrt{29}

\hat{a}=\frac{\:\:}{\sqrt{29} }

Using the relation a=|a|\hat{a}, we have

\:\:=\sqrt{29} \cdot \frac{\:\:}{\sqrt{29} }

c) Let us first find the sum of <1, 2, -3> and <2, 4, 1> to get:

<1+2, 2+4, -3+1>=<3, 6, -2>

Let a=<3, 6, -2>

The magnitude is

|a|=\sqrt{3^2+6^2+(-2)^2}

|a|=\sqrt{9+36+4}=\sqrt{49}=7

The unit vector in the direction of <u>a</u> is

\hat{a}=\frac{\:\:}{7}

Using the relation a=|a|\hat{a}, we have

\:\:=7\cdot \frac{\:\:}{7}

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3 years ago
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