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2 years ago

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Mr. Muñoz has a coupon for 15% off his entire purchase. He buys binoculars for $105 and hiking boots. He spends a total of $170

before tax.
Write and solve an equation to find how much the hiking boots cost, x, before the discount.

1 answer:

True [87]2 years ago

7 0

Answer:

$105(.15)+ x =170$

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mr_godi [17]

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

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

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

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

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

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

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

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

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

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

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

6 0

2 years ago

Kory bought 2 boxes of candy and 1 soda at the movies and paid $8.00.christopher bought 1 box of candy and 1 soda and paid $5.50

viva [34]

Let c represent candy and s soda.
2c+s=8
c+s=5.5
Solving the pair of simultaneous equations by eliminating s,
c=2.5
Representing the value of c in equation 1;
5+s=8
s=3
So, a box of Candy costs 2.5 and a soda costs 3

4 0

2 years ago

Plzzzzzzz help me with #1,2,3

Elodia [21]

Hi once again.
24) D)
25) C) Area of a circle : 3.14× r^2( Radius squared)
26) C)

8 0

2 years ago

Find amplitude and midline of the function y=4cos(4)-5.

Nat2105 [25]

Amplitude: 4

If midline means center, it is -5
I'm also implying that there is an x in cos(4x)

7 0

2 years ago

Questions and answers attached

Black_prince [1.1K]

Answer:

(3)(c) + 43 ≥ 100

Step-by-step explanation:

Given:

Total number of friends = 3 friends

Number of card already have = 43

Total number of card at least want to collect = 100

Find:

Inequality

Computation:

Assume;

Same number of card each friend collect = c

Total number of card at least want to collect ≤ (Total number of friends)(Same number of card each friend collect) + Number of card already have

Total number of card at least want to collect ≤ (3)(c) + 43

100 ≤ (3)(c) + 43

(3)(c) + 43 ≥ 100

3 0

3 years ago