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3 years ago

What is the distance between A (-4,2) and B (12,4)

Mathematics

2 answers:

12345 [234]3 years ago

7 0

Your Answer is (4,3)

My name is Ann [436]3 years ago

3 0

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{2})\qquad
B(\stackrel{x_2}{12}~,~\stackrel{y_2}{4})\qquad \qquad
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AB=\sqrt{[12-(-14)]^2+[4-2]^2}\implies AB=\sqrt{(12+14)^2+(4-2)^2}
\\\\\\
AB=\sqrt{26^2+2^2}\implies AB=\sqrt{680}$

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3 years ago

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A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are

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Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

$P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764$

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

$P = P_{1} + P_{2} + P_{3}$

$P_{1}$ is the probability that all three of them are 13-watt. So:

$P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277$

$P_{2}$ is the probability that all three of them are 18-watt. So:

$P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415$

$P_{3}$ is the probability that all three of them are 23-watt. So:

$P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173$

$P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865$

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

$P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245$

There is a 12.45% probability that one bulb of each type is selected.

3 0

3 years ago

Helppp meeeeeeeeeeeeeee

Leni [432]

Measuring the figure with a ruler, length = 2 inch and width = 1.5 inch

It is given that 1/2 inch = 4 feet

1 inch = 4*2 = 8 feet

Area of rectangle = length * width

Length = 2*8 = 16 feet

Width = 1.5*8 = 12 feet

Area of Quintoe's bedroom = 16 feet * 12 feet = $192 feet^2$

Hence option C is correct.

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3 years ago

Hey can you please help me posted picture of question

Ad libitum [116K]

The correct answer is option C.

Step by step solution is listed below.

$(4 x^{3}+8x)( x^{2} -1) \\ \\ 
 =4 x^{3}(x^{2} -1)+8x(x^{2} -1) \\ \\ 
=4x^{5}-4 x^{3}+8 x^{3}-8x \\ \\ 
=4 x^{5}+4 x^{3}-8x$

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3 years ago

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SVETLANKA909090 [29]

The answer is 35 degree

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3 years ago

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