In ΔHIJ, h = 40 cm, ∠J=20° and ∠H=93°. Find the length of j, to the nearest centimeter.

Answers

Answer 1

Given:

In ΔHIJ, h = 40 cm, ∠J=20° and ∠H=93°.

To find:

The length of j, to the nearest centimeter.

Solution:

According to Law of sine,

[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]

In ΔHIJ, using law of sine, we get

[tex]\dfrac{j}{\sin J}=\dfrac{h}{\sin H}[/tex]

[tex]\dfrac{j}{\sin (20^\circ)}=\dfrac{40}{\sin (93^\circ)}}[/tex]

[tex]j=\dfrac{40\times \sin (20^\circ)}{\sin (93^\circ)}}[/tex]

On further simplification, we get

[tex]j=\dfrac{40\times 0.34202}{0.99863}[/tex]

[tex]j=\dfrac{13.6808}{0.99863}[/tex]

[tex]j=13.69958[/tex]

Approximate the value to the nearest centimeter.

[tex]j\approx 14[/tex]

Therefore, the length of j is 14 cm.


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3. Solve the following application of a system of linear equations.
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Ruben hasn't started yet so he has none. Since Cindy makes 9 flashcards per minute and Ruben makes
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Answers

Answer:

Equations:

[tex]y = 14 + 9 x[/tex] --- Cindy

[tex]y = 10 x[/tex] --- Ruben

Solution to equation:

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Number of cards they have at that time: 140 flashcards

Step-by-step explanation:

Solving (a): Variables and what they represent

The variables to use are x and y

Where x represent the minutes and y represents the number of flashcards in x minutes

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Cindy:

[tex]Existing\ Flashcards = 14[/tex]

[tex]Additional = 9[/tex] per minute

Total number of flashcards (y) in x minutes is:

[tex]y = Existing\ Flashcards + Additional * x[/tex]

[tex]y = 14 + 9 * x[/tex]

[tex]y = 14 + 9 x[/tex]

Ruben:

[tex]Rate = 10[/tex] per minute

Total number of flashcards (y) in x minutes is:

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[tex]y = 10 x[/tex]

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or

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[tex]y = 140[/tex]

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