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Rainbow [258]
2 years ago
15

Pleaseeee need help with this asappp

Mathematics
1 answer:
DerKrebs [107]2 years ago
7 0

Here , we are given with a circle on cartesian plane , and we need to find the <em>equation of the circle </em>, We are also given that it's centre is at <em>(5,-6) </em>{ From graph } and it's radius is<em> 3 units</em> . Now , as we know, the standard equation of a circle is {\bf (x-h)^{2}+(y-k)^{2}=r^{2}} where <em>(h,k) </em>is the centre of the circle and radius is <em>r</em> . Now , our equation will be ;

{:\implies \quad \sf (x-5)^{2}+\{y-(-6)\}^{2}=3^{2}}

{:\implies \quad \bf \therefore \quad \underline{\underline{(x-5)^{2}+(y+6)^{2}=9}}}

<em>This is the required equation of Circle </em>

Or if you want to proceed it further , then you will get <em>x² + y² - 10x + 12y + 50 = 0 </em>

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If you flip the first triangle right side up, we can see that they are congruent in the same exact spots. The two sides are congruent, as well as the angle.

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In triangle JKL, sin(b°) = three fifths and cos(b°) = four fifths. If triangle JKL is dilated by a scale factor of 2, what is ta
scoundrel [369]

Answer:

Tan(b°) = 3/4 = three fourths (C)

Step-by-step explanation:

Triangle JKL is a right angled triangle in which angle K is a right angle and angle L= b°

We would apply SOHCAHTOA from trigonometry to determine the value for each sides.

For ∆JKL

Sin(b°) = opposite/hypotenuse

Sin(b°) = 3/5

Cos(b°) = adjacent/hypotenuse

Cos(b°) = 4/5

Tan(b°) = 3/4

From the above we know the value of each sides of ∆JKL

Find attached the diagram of the above triangle (1)

∆JKL is dilated by a scale factor = 2

Meaning we would multiply each sides of ∆JKL by 2. The values of the new triangle become:

Opposite = 2(3) = 6

Adjacent = 2(4) = 8

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To find tan(b°) of the new triangle, we would apply tangent ratio

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Tan(b°) = 6/8

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Find attached the diagram for the new triangle (2)

Diagram 3 shows the drawing of both triangles together.

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