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vovangra [49]
2 years ago
10

The equation of a circuits in the form: (in the picture)

Mathematics
1 answer:
nexus9112 [7]2 years ago
7 0

Answer: h>0 and k>0

Step-by-step explanation:

If the circle is centered in Quadrant I, then both the x and y coordinates of the center are positive.

This means that h>0 and k>0.

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Can someone plz help me with dis i will give brainliest
Reil [10]

Answer:

first one:  B and C

Second one: x=12, y=10 so the answer is B.

Step-by-step explanation:

First one: Irrational numbers are any numbers that go on forever in the decimal without ever repeating. Any square roots of numbers that aren't perfectly square, such as 4 and 9, are irrational. And anything with pi, or \pi, is always irrational.

Second one: 144 and 100 are perfect squares, meaning it can be created by multiplying the same number by itself, in this case 12 and 10. Subtract those and you get 2.

8 0
3 years ago
Read 2 more answers
In order to answer the following question, please use the following image down below:
Novosadov [1.4K]

Answer:

23

Step-by-step explanation:

(x-3)(9) = (10)(18)

9x - 27 = 180

9x = 207

x = 23

5 0
3 years ago
Given this unit circle, what is the value of x? (x, -7/10)(1,0)​
Paraphin [41]

Answer:

x = - \frac{\sqrt{51} }{10}

Step-by-step explanation:

the equation of a circle centred at the origin is

x² + y² = r² ( r is the radius )

The radius of a unit circle is r = 1

substitute (x, - \frac{7}{10} ) into the equation and solve for x

x² + (- \frac{7}{10} )² = 1²

x² + \frac{49}{100} = 1 ( subtract \frac{49}{100} from both sides )

x² = 1 - \frac{49}{100} = \frac{51}{100} ( take square root of both sides )

x = ± \sqrt{\frac{51}{100} } = ± \frac{\sqrt{51} }{10}

since the point is in the 3rd quadrant then x < 0

x = - \frac{\sqrt{51} }{10}

5 0
2 years ago
Limitation of -4 to the left of the absolute value of x+4 divided by x+4
timama [110]

I believe you're asking about the one-sided limit,

\displaystyle \lim_{x\to-4^-}\frac{|x+4|}{x+4}

Recall the definition of absolute value:

• |x| = x if x\ge0

• |x| = -x if x < 0

Since we're approaching -4 from the left, we're effectively focusing on a domain of x or x+4. So, by the definition above, we have |x+4| = -(x+4). Then in the limit, we have

\displaystyle \lim_{x\to-4^-}\frac{|x+4|}{x+4} = \lim_{x\to-4^-}\frac{-(x+4)}{x+4} = \lim_{x\to-4^-}(-1) = \boxed{-1}

7 0
2 years ago
Henry draws an obtuse triangle how many obtuse angles does Henry's triangle have?
drek231 [11]

it would only have 1 obtuse angle

4 0
3 years ago
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