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34kurt
3 years ago
14

Keisha and David each found the same value for cosine theta, as shown below, given Sine theta = Negative StartFraction 8 Over 17

EndFraction. Keisha’s Solution David’s Solution Tangent squared theta + 1 = secant squared theta. StartFraction sine squared theta Over cosine squared theta EndFraction + 1 = StartFraction 1 Over cosine squared theta EndFraction. StartFraction (eight-seventeenths) squared Over cosine squared theta EndFraction + 1 = StartFraction 1 Over cosine squared theta EndFraction. (eight-seventeenths) squared + cosine squared theta = 1. cosine theta = plus-or-minus StartRoot 1 minus StartFraction 64 Over 289 EndFraction EndRoot. cosine theta = plus-or-minus Fifteen-seventeenths sine squared theta + cosine squared theta = 1. cosine squared theta = 1 minus (negative eight-seventeenths) squared. cosine theta = plus-or-minus StartRoot StartFraction 225 Over 289 EndFraction EndRoot. Cosine theta = plus-or-minus fifteen-seventeenths Whose procedure is correct?
Mathematics
2 answers:
makvit [3.9K]3 years ago
8 0

Answer:

C. Both procedures are correct

e2020

MakcuM [25]3 years ago
4 0

Answer

They are both correct

Step-by-step explanation:

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It would help a lot, and if able, please explain
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2 years ago
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Solve for y 2x+y-3z=5
Elza [17]

Answer:

<h3>A. y=-2x+3z+25</h3>

Step-by-step explanation:

Isolate the term of x and y from one side of the equation.

<u>To solve:</u>

  • The value of y.

\Longrightarrow: \sf{\sqrt{2x+y-3z}=5}

<h3>2x+y-3z=25</h3>

<u>First, you have to subtract by 2x-3z from both sides.</u>

\Longrightarrow: \sf{2x+y-3z-\left(2x-3z\right)=25-\left(2x-3z\right)}}

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\Longrightarrow: \boxed{\sf{y=25-2x+3z}}}

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