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3 years ago

What equation represents a circle whose center is located at (-6,2) and whose radius is 10 units?

1 answer:

3 0

One of the equations for a circle is (x - h)² + (y - k)² = r² where (h, k) are the center and r is the radius.
So after substitution:
(x + 6)² + (y - 2)² = 100

Notice the equation has the opposite of h and k and 100 is the result of squaring 10.

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Find the product. (-4·3·2)2 -48 48 -576 576

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3 years ago

If lines Joined each given point on the graph to the origin, which points would be on lines that represent a unit rate greater t

nirvana33 [79]

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Gnoma [55]

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3 years ago

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In the figure below, CD bisects m∠ACB, AB=BC, m∠BEC=90°, and m∠DCE=42°. Find the measure of ∠A.

vesna_86 [32]

Answer: ∠A= $24\textdegree$

Step-by-step explanation:

Since we have given that

CD bisects m∠ACB,

⇒ m∠ACD= m∠DCB =x (say)

And, AB=BC, m∠BEC=90°, and m∠DCE=42°

So,

m∠CAD=m∠ACD

In ΔDCE,

$\angle DCE+\angle CED+\angle CDE=180\textdegree\\\\(\text{Sum of anlges in triangle is }180\textdegree)\\\\42\textdegree+90\textdegree+\angle CDE=180\textdegree\\\\132\textdegree+\angle CDE=180\textdegree\\\\\angle CDE=180\textdegree-132\textdegree\\\\+\angle CDE=48\textdegree$

Now, in ΔACD,

$x+x=48\textdegree\\\\2x=48\textdegree\\\\\text{( Sum of interior anlges is equal to exterior angles )}\\\\x=\frac{48}{2}\\\\x=24\textdegree$

Hence , ∠A= $24\textdegree$

8 0

3 years ago

Read 2 more answers

Point M is the midpoint of segment QR. If QM = 16 + x and MR = 2(x + 2), find the length of QR. QR = 12 QR = -12 QR = 28 QR = 56

vfiekz [6]

The midpoint divided a segment into two congruent lengths/segments.

So, QM = MR.

Set up the equation.

16 + x = 2(x+ 2)

16 + x = 2x + 4

12 + x = 2x

12 = x

Now, that you have the value of x, substitute it for the two equations.

QM = 16 + x

QM = 16 + 12

QM = 28

MR = 2(12+ 2)

MR = 2(14)

MR = 28

Find the whole length by of the line by adding the two segments. 28 × 2 = 56.

QR = 56

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4 years ago