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3 years ago

Hi kids how are ya ......................................

Mathematics

1 answer:

Bond [772]3 years ago

4 0

Answer:

good

Step-by-step explanation:

i'm good because i'm good

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5. Show that the following points are collinear. a) (1, 2), (4, 5), (8,9)

Irina-Kira [14]

Label the points A,B,C

A = (1,2)
B = (4,5)
C = (8,9)

Let's find the distance from A to B, aka find the length of segment AB.

We use the distance formula.

$A = (x_1,y_1) = (1,2) \text{ and } B = (x_2, y_2) = (4,5)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-4)^2 + (2-5)^2}\\\\d = \sqrt{(-3)^2 + (-3)^2}\\\\d = \sqrt{9 + 9}\\\\d = \sqrt{18}\\\\d = \sqrt{9*2}\\\\d = \sqrt{9}*\sqrt{2}\\\\d = 3\sqrt{2}\\\\$

Segment AB is exactly $3\sqrt{2}$ units long.

Now let's find the distance from B to C

$B = (x_1,y_1) = (4,5) \text{ and } C = (x_2, y_2) = (8,9)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(4-8)^2 + (5-9)^2}\\\\d = \sqrt{(-4)^2 + (-4)^2}\\\\d = \sqrt{16 + 16}\\\\d = \sqrt{32}\\\\d = \sqrt{16*2}\\\\d = \sqrt{16}*\sqrt{2}\\\\d = 4\sqrt{2}\\\\$

Segment BC is exactly $4\sqrt{2}$ units long.

Adding these segments gives

$AB+BC = 3\sqrt{2}+4\sqrt{2} = 7\sqrt{2}$

----------------------

Now if A,B,C are collinear then AB+BC should get the length of AC.

AB+BC = AC

Let's calculate the distance from A to C

$A = (x_1,y_1) = (1,2) \text{ and } C = (x_2, y_2) = (8,9)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-8)^2 + (2-9)^2}\\\\d = \sqrt{(-7)^2 + (-7)^2}\\\\d = \sqrt{49 + 49}\\\\d = \sqrt{98}\\\\d = \sqrt{49*2}\\\\d = \sqrt{49}*\sqrt{2}\\\\d = 7\sqrt{2}\\\\$

AC is exactly $7\sqrt{2}$ units long.

Therefore, we've shown that AB+BC = AC is a true equation.

This proves that A,B,C are collinear.

For more information, check out the segment addition postulate.

7 0

2 years ago

The orbital period, P, of a planet and the planet’s distance from the sun, a, in astronomical units is related by the formula P

mixas84 [53]

Answer:

9.54731702 astronomical units

Step-by-step explanation:

$P=a^{\frac{3}{2} }$

Substitute this value in the equation above:

$29.5=a^{\frac{3}{2} }$

⇒ $(29.5)^{\frac{3}{2} } =a$

⇒ $a=9.54731702$ $astronomical$ $units$

Hence, its distance from the sun is 9.54731702 astronomical units.

7 0

3 years ago

F(x) = -2|x-2|+4 vertical stretch by a factor of 2

stepladder [879]

A vertical stretch by a factor of 2 of F(x) = -2|x-2|+4 would result in the revised function G(x) = -4|x-2|+4. The vertex would stay in the same place (2,4).

5 0

3 years ago

Find the length of ab

dezoksy [38]

I think it’s two, counting the lines

7 0

3 years ago

Read 2 more answers

Eric and Sue will randomly select from a treat bag containing 3 lollipops and 2 gum balls. Eric will select, replace it, and the

Vilka [71]

Answer:

Sue has the greater probability of 2/5 and Eric has the probability of 1/5.

Step-by-step explanation:

Sue took 2 things and eric replaced his forst one, ending up with one treat.

8 0

3 years ago