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2 years ago

14

A candy store sells caramels for $9/pound and chocolate for $12/pound. How much more is the cost of 2 pounds of caramel than 1.5

pounds of chocolate? *Show your work*

1 answer:

Rashid [163]2 years ago

5 0

Answer:

The caramel is $0 more than the chocolate. In other words, they cost the same.

Step-by-step explanation:

Since we have 2 pounds of caramel and caramel costs $9/pound, the cost of 2 pounds of caramel can be found by this equation:

2lbs · $9 = $18

Since we have 1.5 pounds of chocolate and chocolate costs $12/pound, the cost of 1.5 pounds of chocolate can be found by this equation:

1.5lbs · $12 = $18

To find how much bigger the cost of caramel is than that of chocolate, we will subtract the total cost of the chocolate from the total cost of the caramel.

$18 - $18 = $0

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Complete the assignment on a separate sheet of paper<br><br> Please attach pictures of your work.

Irina18 [472]

Answer:

<u>TO FIND :-</u>

Length of all missing sides.

<u>FORMULAES TO KNOW BEFORE SOLVING :-</u>

$\sin \theta = \frac{Side \: opposite \: to \: \theta}{Hypotenuse}$
$\cos \theta = \frac{Side \: adjacent \: to \: \theta}{Hypotenuse}$
$\tan \theta = \frac{Side \: opposite \: to \: \theta}{Side \: adjacent \: to \: \theta}$

<u>SOLUTION :-</u>

1) θ = 16°

Length of side opposite to θ = 7

Hypotenuse = x

$=> \sin 16 = \frac{7}{x}$

$=> \frac{7}{x} = 0.27563......$

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2) θ = 29°

Length of side opposite to θ = 6

Hypotenuse = x

$=> \sin 29 = \frac{6}{x}$

$=> \frac{6}{x} = 0.48480......$

$=> x = \frac{6}{0.48480....} = 12.37599.....$ ≈ 12.3

3) θ = 30°

Length of side opposite to θ = x

Hypotenuse = 11

$=> \sin 30 = \frac{x}{11}$

$=> \frac{x}{11} = 0.5$

$=> x = 0.5 \times 11 = 5.5$

4) θ = 43°

Length of side adjacent to θ = x

Hypotenuse = 12

$=> \cos 43 = \frac{x}{12}$

$=> \frac{x}{12} = 0.73135......$

$=> x = 12 \times 0.73135.... = 8.77624....$ ≈ 8.8

5) θ = 55°

Length of side adjacent to θ = x

Hypotenuse = 6

$=> \cos 55 = \frac{x}{6}$

$=> \frac{x}{6} = 0.57357......$

$=> x = 6 \times 0.57357.... = 3.44145....$ ≈ 3.4

6) θ = 73°

Length of side adjacent to θ = 8

Hypotenuse = x

$=> \cos 73 = \frac{8}{x}$

$=> \frac{8}{x} = 0.29237......$

$=> x = \frac{8}{0.29237.....} = 27.36242.....$ ≈ 27.3

7) θ = 69°

Length of side opposite to θ = 12

Length of side adjacent to θ = x

$=> \tan 69 = \frac{12}{x}$

$=> \frac{12}{x} = 2.60508......$

$=> x = \frac{12}{2.60508....} = 4.60636....$ ≈ 4.6

8) θ = 20°

Length of side opposite to θ = 11

Length of side adjacent to θ = x

$=> \tan 20 = \frac{11}{x}$

$=> \frac{11}{x} = 0.36397......$

$=> x = \frac{11}{0.36397....} =30.22225....$ ≈ 30.2

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2 years ago

List all x intercepts for y= -cos(1/2x + 1/5 pi) on the interval [-2/5pi,4pi]

Dahasolnce [82]

- cos ( 1/2 x + 1/5 π ) = 0 ( and because if cos α = 0, α= π/2 + k π, k ∈ Z )
1/2 x + π/5 = π/2 + k π, k ∈ Z
1/2 x = π/2 - π/5 + k π / * 2
x = π - 2π/5 + 2 k π
x = 3/5 π + 2 k π = 0.6 π + 2 k π
Answer:
If k = 0: x 1 = 0.6 π = 3π/5
k = 1 : x 2 = 2.6 π = 13π/5

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