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harina [27]
4 years ago
9

How to find the missing angle

Mathematics
2 answers:
masya89 [10]4 years ago
8 0
You have to add up the two known angles.
Natali5045456 [20]4 years ago
3 0
Sorry i could not understand it clearly make it tidy
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Find the constant difference for a hyperbola with foci (-3.5, 0) and (3.5, 0) and a point on the hyperbola (3. 5, 24).
Solnce55 [7]

Answer:

2a=1

Step-by-step explanation:

The constant difference for a hyperbola \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 is  2a.

1. Since hyperbola has foci (-3.5, 0) and (3.5, 0), then c=3.5. Note that c=\sqrt{a^2+b^2}, then

\sqrt{a^2+b^2}=3.5\Rightarrow a^2+b^2=3.5^2.

2. Since point (3.5,24) lies on the hyperbola, then

\dfrac{3.5^2}{a^2}-\dfrac{24^2}{b^2}=1.

3. Solve the system of two equations:

\left\{\begin{array}{l}a^2+b^2=3.5^2\\\dfrac{3.5^2}{a^2}-\dfrac{24^2}{b^2}=1\end{array}\right.

From the 1st equation,

b^2=3.5^2-a^2,

then

\dfrac{3.5^2}{a^2}-\dfrac{24^2}{3.5^2-a^2}=1,\\ \\\dfrac{3.5^2(3.5^2-a^2)-24^2a^2}{a^2(3.5^2-a^2)}=1,\\ \\3.5^4-3.5^2a^2-24^2a^2=3.5^2a^2-a^4,\\ \\a^4-a^2(2\cdot 3.5^2+24^2)+3.5^4=0,\\ \\a^4-600.5a^2+150.0625=0,\\ \\D=(-600.5)^2-4\cdot 150.0625=360000,\\ \\a^2_{1,2}=\dfrac{600.5\pm 600}{2}=\dfrac{1}{4},600\dfrac{1}{4}.

For a^2=\dfrac{1}{4},\ b^2=3.5^2-0.25=12.

For a^2=600\dfrac{1}{4},\ b^2=3.5^2-600.25 this is impossible, then a^2=600\dfrac{1}{4} is extra solution.

Hence, a=\dfrac{1}{2} and 2a=1.

5 0
3 years ago
The area of a square painting is 81p^2+90p+25 what is the side length of the paining
IgorLugansk [536]
81p²+90p+25=(9p)²+2*9p*5+5²=(9p+5)²
so the side length of the paining is 9p+5


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3 years ago
Which of the following is a defined term?
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What of the following ? You need to add more then define term what are the following?
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I'm putting C but I'm not sure ill let you know if its right after the test

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