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2 years ago

If the endpoints of the diameter of a circle are (−10, −8) and (−6, −2), what is the standard form equation of the circle?

2 answers:

7 0

First, find the midpoint of the diameter. the midpoint is the center of the circle.
[(-10+(-6)]/2=-8, [(-8+(-2)]/2=-5
so the center is at (-8, -5)
next, find the distance between the two points:
d=√[-6-(-10)]²+[-2-(-8)]²=√4²+6²=√52=2√13
half of the diameter is radius, r=half of 2√13=√13
standard equation for the circle is (x-h)²+(y-k)²=r², (h,k) being the center, r the radius
(x+8)²+(y+5)²=13 is the answer.

Mekhanik [1.2K]2 years ago

3 0

Answer:

(x + 8)2 + (y + 5)2 = 13

Step-by-step explanation:

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What is the result when 12a + (-8b) is subtracted from 24a + 10b

Sever21 [200]

Hello from MrBillDoesMath!

Answer:

12a + 18b

Discussion:

The question asks us to calculate

24a + 10 b - (12a - 8b) => (combine common terms)

(24a - 12a) + ( 10b + 8b) => ( as - (-8b) = + 8b)

12a + 18b

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2 years ago

For the data in the table, does y vary directly with x? If it does, write an equation for the direct variation.

QveST [7]

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

X - 35 > 15 i need answer IMEDIATLEY

SIZIF [17.4K]

I'm assuming you want us to solve this inequality for x, since the question wasn't too clear on that.

$x - 35 > 15$

Add 35 to both sides to remove the "-35" on the left side so we can isolate the x

$x-35+35>15+35$

$x>50$

The solution to this inequality is x>50. Let me know if you need any clarifications, thanks!

~ Padoru

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2 years ago

Use a graphic organizer to compare and contrast similar and congruent figures

Anika [276]

Similar figures are figures of the same shape but different (or same) size.
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8 0

2 years ago

Suppose a simple random sample of size nequals81 is obtained from a population with mu equals 79 and sigma equals 18. (a) Descr

Daniel [21]

Answer:

(a) The sampling distribution of $\overline{X}$ = Population mean = 79

(b) P ( $\overline{X}$ greater than 81.2 ) = 0.1357

(c) P ( $\overline{X}$ less than or equals 74.4 ) = .0107

(d) P (77.6 less than $\overline{X}$ less than 83.2 ) = .7401

Step-by-step explanation:

Given -

Sample size ( n ) = 81

Population mean $(\nu)$ = 79

Standard deviation $(\sigma )$ = 18

(a) Describe the sampling distribution of $\overline{X}$

For large sample using central limit theorem

the sampling distribution of $\overline{X}$ = Population mean = 79

(b) What is Upper P ( $\overline{X}$ greater than 81.2 ) =

$P(\overline{X}> 81.2)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}> \frac{81.2 - 79}{\frac{18}{\sqrt{81}}})$

= $P(Z> 1.1)$

= $1 - P(Z< 1.1)$

= 1 - .8643 =

= 0.1357

$P(\overline{X}\leq 74.4)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{74.4- 79}{\frac{18}{\sqrt{81}}})$

= $P(Z\leq -2.3)$

= .0107

(d) What is Upper P (77.6 less than $\overline{X}$ less than 83.2 ) =

$P(77.6< \overline{X}< 83.2)$ = $P(\frac{77.6- 79}{\frac{18}{\sqrt{81}}})< P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{83.2- 79}{\frac{18}{\sqrt{81}}})$

= $P(- 0.7< Z< 2.1)$

= $(Z< 2.1) - (Z< -0.7)$

= 0.9821 - .2420

= 0.7401

3 0

2 years ago