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8090 [49]
3 years ago
8

Need help solving it. Quantitative Literacy.

Mathematics
1 answer:
ioda3 years ago
5 0

1 lb ≈ 0.454 kg

1 in ≈ 2.54 cm

I'm not sure I understand the "do not" part of this question. What does it mean to convert from cubic inches (volume) to inches (length), and from centimeter to cubic centimeters?

In any case, you have

(1 lb)/(32.375 in³) ≈ (1 lb)/(3.19 in)³

… ≈ (1 lb)/(3.19 in)³ × (0.454 kg)/(1 lb) × ((1 in)/(2.54 cm))³

… ≈ (0.454 kg)/(3.19 in)³ × ((1 in)/(2.54 cm))³

… ≈ (0.454 kg) × ((1 in)/((3.19 in) × (2.54 cm))³

… ≈ (0.454 kg) × (1/(8.10 cm))³

… ≈ (0.454 kg) × 1/(530.531 cm³)

… ≈ (0.454 kg)/(530.531 cm³)

… ≈ 0.000856 kg/cm³

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Answer: \text{Length of AG=}\frac{2\sqrt{63}}{3}

Explanation:  

Please follow the diagram in attachment.  

As we know median from vertex C to hypotenuse is CM  

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We are given length of CG=4  

Median divide by centroid 2:1  

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Where, CG=4

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Length of CM=4+2= 6 ft  

\therefore CM=\frac{1}{2}AB\Rightarrow AB=12

In \triangle ABC, \angle C=90^0

Using trigonometry ratio identities  

AC=AB\sin 30^0\Rightarrow AC=6 ft

BC=AB\cos 30^0\Rightarrow BC=6\sqrt{3} ft  

CN=\frac{1}{2}BC\Rightarrow CN=3\sqrt{3} ft

In \triangle CAN, \angle C=90^0  

Using pythagoreous theorem  

AN=\sqrt{6^2+(3\sqrt{3})^2\Rightarrow \sqrt{63}

Length of AG=2/3 AN

\text{Length of AG=}\frac{2\sqrt{63}}{3} ft


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Based on the information given, it can be noted that the final price of the item will be $74500.

<h3>Solving percentages.</h3>

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