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Alexxx [7]
2 years ago
10

One kind of hard candy sells for $.89 per kilogram another sells for $1.10 per kilogram how many kilograms of each kind a need t

o be used for 30 kg of a mixture to sell for $.96 per kilogram
Mathematics
1 answer:
Furkat [3]2 years ago
4 0

Answer:

20kg of $0.89 candy

10kg of $1.10 candy

Step-by-step explanation:

Candy 1 = 0.89 per kg

Candy 2 = 1.10 per kg

Total kilogram, kg = 30

Let candy 1 = x ; candy 2 = (30 - x) ;

0.89x + 1.10(30 - x) = 0.96(30)

0.89x + 33 - 1.10x = 28.8

0.89x - 1.10x = 28.8 - 33

-0.21x = - 4.2

x = 4.2 / 0.21

x = 20

20kg of $0.89 candy

(30 - x) = (30 - 20) = 10kg

10kg of $1.10 candy

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<u>Given</u>:

Given that the figure is a triangular prism.

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<u>Volume of the triangular prism:</u>

The volume of the triangular prism can be determined using the formula,

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3 years ago
A flagpole is located at the edge of a sheer y = 70-ft cliff at the bank of a river of width x = 40 ft. See the figure below. An
Gnom [1K]
I would solve this using tangents.  Let h be height of flagpole.
Set up 2 right triangles, each with a base of 40.
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Note: A-B = 9
To solve for h we need to use the "Difference Angle" formula for Tangent
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Plug in what we know:
tan(9) = \frac{ \frac{h+70}{40} -  \frac{70}{40}}{1+ (\frac{h+70}{40})(\frac{7}{4})}
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