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denis-greek [22]

2 years ago

14

1. A restaurant charges $10 for each burrito, and a $5 delivery fee. Write an expression to represent the total cost to have x b

urritos delivered.
2. What is the total cost to have 4 burritos delivered ?

1 answer:

Andrew [12]2 years ago

3 0

1. y=10x+5

2. 10(4)+5 = 45.

So, 45 is the total cost to have 4 burritos delivered.

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What is A = 8x + 74 degrees and B = 2x + 56 degrees solve for x and then find the measure of B

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Since those lines are parallel, we know that A + B = 180°

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Answer:

$V = \frac{2L\sqrt{3}}{3} \pi}$

$A = 4\pi + \sqrt{3L^{2} + 16}$

Step-by-step explanation:

Figure of cone is missing. See attachment

Given

Radius, R = 2m

Let L = KL=LM=KM

Required:

Volume, V and Surface Area, A

<u><em>Calculating Volume</em></u>

Volume is calculated using the following formula

$V = \frac{1}{3} \pi R^{2} H$

Where R is the radius of the cone and H is the height

First, we need to determine the height of the cone

The height is represented by length OL

It is given that KL=LM=KM in triangle KLM

This means that this triangle is an equilateral triangle

where OM = OK = $\frac{1}{2} KL$

OK = $\frac{1}{2}L$

Applying pythagoras theorem in triangle LOM,

|LM|² = |OL|² + |OM|²

By substitution

L² = H² + ( $\frac{1}{2}L$ )²

H² = L² - $\frac{1}{4}L$ ²

H² = L² (1 - $\frac{1}{4}$ )

H² = L² $\frac{3}{4}$

H² = $\frac{3L^{2} }{4}$

Take square root of bot sides

$H = \sqrt{\frac{3L^{2} }{4}}$

$H = \frac{L\sqrt{3}}{2}$

Recall that $V = \frac{1}{3} \pi R^{2} H$

$V = \frac{1}{3} \pi 2^{2} * \frac{L\sqrt{3}}{2}$

$V = \frac{1}{3} \pi * 4} * \frac{L\sqrt{3}}{2}$

$V = \frac{1}{3} \pi} * {2L\sqrt{3}}$

$V = \frac{2L\sqrt{3}}{3} \pi}$

in terms of $\pi$ an d L where L = KL = LM = KM

<u><em>Calculating Surface Area</em></u>

Surface Area is calculated using the following formula

$H = \frac{L\sqrt{3}}{4}$

$A=\pi r(r+\sqrt{h^{2} +r^{2} } )$

$A=\pi * 2(2+\sqrt{((\frac{L\sqrt{3}}{2})^{2} +2^{2} } )$ )

$A=\pi * 2(2+\sqrt{{\frac{3L^{2}}{4} } + 4 }$ )

$A=\pi * 2(2+\sqrt{{\frac{3L^{2}+16}{4} } })$

$A=2\pi(2+\sqrt{{\frac{3L^{2}+16}{4} } })$

$A = 2\pi (2 + \frac{\sqrt{3L^{2} + 16}}{\sqrt{4}} )$

$A = 2\pi (2 + \frac{\sqrt{3L^{2} + 16}}{2} )$

$A = 2\pi (2 + {\frac{1}{2} \sqrt{3L^{2} + 16})$

$A = 4\pi + \sqrt{3L^{2} + 16}$

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