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2 years ago

12

The measures of the angles of a triangle are shown in the figure below. Solve for x. 57°

2 answers:

lorasvet [3.4K]2 years ago

8 0

Answer:

<em>x</em> = 33°

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u> </u>

<u>Geometry</u>

All angles in a triangle add up to 180°

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify and set up</em>

x° + 90° + 57° = 180°

<u>Step 2: Solve for </u><em><u>x</u></em>

[Addition] Combine like terms: x° + 147° = 180°
[Subtraction Property of Equality] Subtract 147 on both sides: x° = 33°

denis-greek [22]2 years ago

4 0

Answer:

Angle = 33

Step-by-step explanation:

Angles should add up to 180 and the angles shown are 90 and 57 and then add those to get 147 then subtract 147 by 180 to get 33 for the angle. Hope this helps!

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Klio2033 [76]

First part of question:

Find the general term that represents the situation in terms of k.

The general term for geometric series is:

$a_{n}=a_{1}r^{n-1}$

$a_{1}$ = the first term of the series

$r$ = the geometric ratio

$a_{1}$ would represent the height at which the ball is first dropped. Therefore:

$a_{1} = k$

We also know that the ball has a rebound ratio of 75%, meaning that the ball only bounces 75% of its original height every time it bounces. This appears to be our geometric ratio. Therefore:

$r=\frac{3}{4}$

Our general term would be:

$a_{n}=a_{1}r^{n-1}$

$a_{n}=k(\frac{3}{4}) ^{n-1}$

Second part of question:

If the ball dropped from a height of 235ft, determine the highest height achieved by the ball after six bounces.

$k$ represents the initial height:

$k = 235\ ft$

$n$ represents the number of times the ball bounces:

$n = 6$

Plugging this back into our general term of the geometric series:

$a_{n}=k(\frac{3}{4}) ^{n-1}$

$a_{n}=235(\frac{3}{4}) ^{6-1}$

$a_{n}=235(\frac{3}{4}) ^{5}$

$a_{n}=55.8\ ft$

$a_{n}$ represents the highest height of the ball after 6 bounces.

Third part of question:

If the ball dropped from a height of 235ft, find the total distance traveled by the ball when it strikes the ground for the 12th time. 

This would be easier to solve if we have a general term for the <em>sum </em>of a geometric series, which is:

$S_{n}=\frac{a_{1}(1-r^{n})}{1-r}$

We already know these variables:

$a_{1}= k = 235\ ft$

$r=\frac{3}{4}$

$n = 12$

Therefore:

$S_{n}=\frac{(235)(1-\frac{3}{4} ^{12})}{1-\frac{3}{4} }$

$S_{n}=\frac{(235)(1-\frac{3}{4} ^{12})}{\frac{1}{4} }$

$S_{n}=(4)(235)(1-\frac{3}{4} ^{12})$

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Answer:

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Step-by-step explanation:

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So we can plug in 14 for b to find c.

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Now this equation can be factored into:

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