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2 years ago

Factorización de 6y-12x2+18x3

Mathematics

2 answers:

givi [52]2 years ago

8 0

Factor- 6(y+5) BOOM THATS THE ANSWER :D

wariber [46]2 years ago

5 0

Answer:

Factor- 6(y+5)

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Someone please help me with number ten!!

ivann1987 [24]

Answer:

Step-by-step explanation:

starts with 4 and is multiplied by 3 every term

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2 years ago

Jada asked some students at her school how many hours they spent watching television last week, to the nearest hour. Here are a

-Dominant- [34]

Answer:

100

Step-by-step explanation:

Based on the histogram given :

The vertical axis gives the number of student in the school :

For the range :

0 - 5 = 40

5 - 10 = 30

10 - 15 = 20

15 - 20 = 5

20 - 25 = 3

25 - 30 = 2

Taking the sum :

(40 + 30 + 20 + 5 + 3 + 2) = 100

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2 years ago

Pamela took a total of 27 quizzes over the course of 9 weeks. After attending 10 weeks of school this quarter, how many quizzes

shepuryov [24]

Answer:

Step-by-step explanation:

27 diviend by 9 = 3 so 27 +3 = 30.

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2 years ago

If one line has a slope of 2/3, then the slope of a line parallel to it would be?

ipn [44]

-3/2

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3 years ago

On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th

Elena L [17]

Answer:

The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$

Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, a = $\sqrt{(3-0)^{2}+(4-0)^{2}}$

Or, a = $\sqrt{9+16}$

Or, a = $\sqrt{25}$

∴ a = 5 unit

Similarly

The measure of length BC = b = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, b = $\sqrt{(3-3)^{2}+(2-4)^{2}}$

Or, a = $\sqrt{0+4}$

Or, b = $\sqrt{4}$

∴ b = 2 unit

And

So, The measure of length CA = c = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, c = $\sqrt{(3-0)^{2}+(2-0)^{2}}$

Or, c = $\sqrt{9+4}$

Or, c = $\sqrt{13}$

∴ c = $\sqrt{13}$ unit

Now, area of Triangle written as , from Heron's formula

A = $\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$

and s = $\frac{a+b+c}{2}$

I.e s = $\frac{5+2+\sqrt{13}}{2}$

Or. s = $\frac{7+\sqrt{13}}{2}$

So, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times ((\frac{(7+\sqrt{13})}{2})-5)\times (\frac{7+\sqrt{13}}{2}-2)\times (\frac{7+\sqrt{13}}{2}-\sqrt{13})}$

Or, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times (\frac{(\sqrt{13}-3)}{2})\times (\frac{4+\sqrt{13}}{2})\times (\frac{7-\sqrt{13}}{2})}$

Or, A = $\frac{3}{2}$ × $\sqrt{1+\sqrt{13} }$

∴ Area of triangle = 1.5 $\sqrt{4.6}$

Hence The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$ Answer

7 0

2 years ago

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