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3 years ago

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An ice cream cone has a radius of 3 inches and a height of 9 inches. What is the exact value of the volume of the ice cream cone

?
81π in3

27π in3

9π in3

3π in3

2 answers:

garri49 [273]3 years ago

4 0

The equation used to find the equation of a cone is ×radius²×height/3

Find your dimensions so it is easy to plug them into the equation:

radius: 3 in

height: 9 in

Plug these values into your equation: ×(3)²×9/3

The volume of the cone is 27π in3

katrin2010 [14]3 years ago

3 0

Answer: I believe its 27π in3

Step-by-step explanation: good luck

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2 years ago

Ryan and Sarah collect baseball cards. Ryan has 150 baseball cards and collects 10 cards per week. Sarah has 200 baseball cards

Nadusha1986 [10]

Answer:

<h2>10 weeks </h2>

Step-by-step explanation:

Step one:

given data

Ryan

number of baseball cards=150

number collected per week= 10

let the number of weeks be x

and the total be y

y=10x+150-----------------1

Sarah

number of baseball cards=200

number collected per week= 5

let the number of weeks be x

and the total be y

y=5x+200------------2

Step two:

Required

the number of weeks where both total will be the same

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3 years ago

Determine the number of significant digits in the measurement <br> 0.79 cm

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3 years ago

Х- а<br>x-b<br>If f(x) = b.x-a÷b-a + a.x-b÷a - b<br>Prove that: f (a) + f(b) = f (a + b)

GenaCL600 [577]

Given:

Consider the given function:

$f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot(x-b)}{a-b}$

To prove:

$f(a)+f(b)=f(a+b)$

Solution:

We have,

$f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot (x-b)}{a-b}$

Substituting $x=a$ , we get

$f(a)=\dfrac{b\cdot(a-a)}{b-a}+\dfrac{a\cdot (a-b)}{a-b}$

$f(a)=\dfrac{b\cdot 0}{b-a}+\dfrac{a}{1}$

$f(a)=0+a$

$f(a)=a$

Substituting $x=b$ , we get

$f(b)=\dfrac{b\cdot(b-a)}{b-a}+\dfrac{a\cdot (b-b)}{a-b}$

$f(b)=\dfrac{b}{1}+\dfrac{a\cdot 0}{a-b}$

$f(b)=b+0$

$f(b)=b$

Substituting $x=a+b$ , we get

$f(a+b)=\dfrac{b\cdot(a+b-a)}{b-a}+\dfrac{a\cdot (a+b-b)}{a-b}$

$f(a+b)=\dfrac{b\cdot (b)}{b-a}+\dfrac{a\cdot (a)}{-(b-a)}$

$f(a+b)=\dfrac{b^2}{b-a}-\dfrac{a^2}{b-a}$

$f(a+b)=\dfrac{b^2-a^2}{b-a}$

Using the algebraic formula, we get

$f(a+b)=\dfrac{(b-a)(b+a)}{b-a}$ $[\because b^2-a^2=(b-a)(b+a)]$

$f(a+b)=b+a$

$f(a+b)=a+b$ [Commutative property of addition]

Now,

$LHS=f(a)+f(b)$

$LHS=a+b$

$LHS=f(a+b)$

$LHS=RHS$

Hence proved.

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3 years ago