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Gala2k [10]
1 year ago
13

Identify the equation of the translated graph in general form x^2-y^2=9 for T(-4,2)

Mathematics
2 answers:
Lelechka [254]1 year ago
8 0

Answer:

B

Step-by-step explanation:

The equation x^2-y^2=a under the translation of  T (p,q) will have the form  (x-p)^2-(y-q)^2=a

<em>Now, using the translation T(-4,2) in the equation given, we can write:</em>

<em>(x+4)^2-(y-2)^2-9=0\\x^2+8x+16-y^2+4y-4-9=0\\x^2-y^2+8x+4y+3=0</em>

<em />

<em>Looking at the choices, </em><em>B is the right answer.</em>

liberstina [14]1 year ago
7 0

Answer:

Looking at the choices, B is the right answer.

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Determine the value of $a$. [asy] pair w=(0,4); pair x=(0,0); pair y=(4,0); pair z=y+7/sqrt(2)*(1,1); dot(w); dot(x); dot(y); do
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Answer:

a=7

Step-by-step explanation:

The image is rendered and attached below.

Triangle WXY is an Isosceles right triangle, since WX=XY.

First, we determine the length of WY using Pythagoras Theorem.

WY=\sqrt{4^2+4^2}\\WY=\sqrt{32}

Since triangle WXY is Isosceles, \angle XYW=45^\circ

\angle XYZ=\angle XYW+\angle WYZ\\135^\circ=45^\circ+\angle WYZ\\\angle WYZ=135^\circ-45^\circ=90^\circ

Therefore:

Triangle WYZ is a right triangle with WZ as the hypothenuse.

Applying Pythagoras Theorem

WZ^2=WY^2+YZ^2\\9^2=(\sqrt{32})^2+a^2\\a^2=81-32\\a^2=49\\a^2=7^2\\$Therefore: a=7

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Evaluate 4-0.25g+0.5h4−0.25g+0.5h4, minus, 0, point, 25, g, plus, 0, point, 5, h when g=10g=10g, equals, 10 and h=5h=5h, equals,
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I believe the correct given equation is in the form of:

4 – 0.25 g + 0.5 h

Now we are to evaluate the equation with the given values:

g = 10 and h = 5

What this actually means is that to evaluate simply means to calculate for the value of the equation by plugging in the values of the variables. Therefore:

4 – 0.25 g + 0.5 h = 4 – 0.25 (10) + 0.5 (5)

4 – 0.25 g + 0.5 h = 4 – 2.5 + 2.5

4 – 0.25 g + 0.5 h = 4

 

Therefore the value of the equation is:

4

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Answer:

The answer is 72 degrees

Step-by-step explanation:

The picture that Helpmetnx showed does work. But they made a mistake and assumed that the diagonal is a angle bisector, and it's not.

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