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AveGali [126]
3 years ago
12

What is the area of a triangle with b=146.2, c=209.3, and A=62.2°?

Mathematics
1 answer:
kotykmax [81]3 years ago
3 0

Answer:

Your answer would be d.

Step-by-step explanation:

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The vertices of a hyperbola are located at (−4, 1) and (4, 1). The foci of the same hyperbola are located at (−5, 1) and (5, 1).
topjm [15]

Answer:

x^2\ 16 - y^2/9 = 1

step-by-step explanation:

soooo the equation for a hyperbola that is horizontal is x^2/a^2 - y^2/b^2 = 1

a hyperbola should always equal one so dont forget that when writing your equation becuase it is easy to forget.

it helps to graph this so you can see it better

to find a it is the distance from the center to the vertices which is 4 so in the equation you will write 16 becuase it is a^2

then you need to find b. to get b you have to figure out that c is the distance from the center to the foci which is 5 and it is all related to the plythagorm theorum becuase it forms a right triangle. so you do c^2 - a^2 = b^2

you get 9 for b^2 because 25-16=9 and so you put that in the equation

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3 years ago
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Find g(4x)<br> g(x)=x²-4
7nadin3 [17]

Answer:

g(x) = x² - 4 is already in form of a variable, I.e., x

g(4x) takes another variable, I.e., 4x

Same as before, 4x takes over x:

=> g(4x) = (4x)² - 4

  • <em>(</em><em>ax</em><em>)</em><em>²</em><em> </em><em>=</em><em> </em><em>a</em><em>²</em><em>x</em><em>²</em><em>,</em><em> </em><em>where</em><em> </em><em>a</em><em> </em><em>is</em><em> </em><em>some</em><em> </em><em>arbitrary</em><em> </em><em>constant</em><em>.</em><em> </em>

<h3><u>Answer</u><u>:</u> </h3>

=> g(4x) = 16x² - 4

OR

=> g(4x) = 4{4x² - 1}

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