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3 years ago

The vertices of a triangle are A (-2, 2), B (4,10), and C (0, -2). Find the equation of the line that contains

Mathematics

1 answer:

Tpy6a [65]3 years ago

4 0

Answer:

y = 1/2 x + 3

Step-by-step explanation:

mid point of line BC = (2, 4)

slope of the line containing median (m) = 1/2

y intercept (b)= 3

since y = mx + b

y = 1/2x + 3

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4. Given the points M(-3, -4) and T(5,0), find the coordinates of the point Q on directed line segment MT that partitions MT in

Firlakuza [10]

Answer:

4 :  Coordinate of Q = $(\frac{1}{5} ,\frac{-12}{5} )$

5 : Coordinate of point A = (-2,-2)

Explanation :

4) Given the points M(-3, -4) and T(5,0)

x₁= -3, y₁ = -4 and

x₂ = 5, y₂ = 0

m = 2, n = 1

Now apply the section formula,

{(mx₂+nx₁)/(m+n) , (my₂+ny₁)/(m+n)}

Coordinate of point Q = $(\frac{1}{5} ,\frac{-12}{5} )$

5) AB:BC = 3:4

A = (x,y), end point

B = (4, 1)

C = (12, 5) end point

So, m = 3 & n = 4

x₁ = x , y₁ = y , x₂=12 & y₂ = 5

Apply section formula we get

4 = (36+4x)/7 & 1 = (15 + 4y)/7

28= 36 + 4x 7 = 4y + 15

4x =-8 4y =-8

x = -2 y = -2

5 0

3 years ago

Solve. 3(x + 1) - 2x = -6 * x = 1 x = 5 x = -7 x = -9

Vadim26 [7]

The answer is x=9 because if you do 3x+-2 that gives you 1 and 1 divided by 9 is 9 and you get that by getting rid of your 6 first and adding it to the 3 from 3(x+1)

7 0

3 years ago

If 2 > -a, then _____. a > 2 a > -2 a < 2 a < -2

FinnZ [79.3K]

Answer:

B. a >-2

Step-by-step explanation:

Hope this helps! Ask me anything if you have any quistions!

4 0

3 years ago

Read 2 more answers

True or False. i think this is easy but

inna [77]

Answer:

of course its true

Step-by-step explanation:

8 0

3 years ago

HEEELP

Drupady [299]

Answer: (c)

Step-by-step explanation:

Given

$f(x)=\dfrac{1}{x-3}\\\\g(x)=\sqrt{x+5}$

Here, $\sqrt{x+5}\ \text{is always greater than equal to 0}\\\Rightarrow x+5\geq 0\\\Rightarrow x\geq -5\quad \ldots(i)$

To get $f\left(g(x)\right)$ , replace $x$ in $f(x)$ by $g(x)\ \text{i.e. by}\ \sqrt{x+5}$

$\Rightarrow f\left(g(x)\right)=\dfrac{1}{\sqrt{x+5}-3}\\\\\text{Denominator must not be equal to 0}\\\\\therefore \sqrt{x+5}-3\neq0\\\Rightarrow \sqrt{x+5}\neq 3\\\Rightarrow x+5\neq 9\\\Rightarrow x\neq 4\quad \ldots(ii)$

Using $(i)$ and $(ii)$ it can be concluded that the domain of $f\left(g(x)\right)$ is all real numbers except 0.

Therefore, its domain is given by

$x\in [-5,4)\cup (4,\infty)$

Option (c) is correct.

5 0

2 years ago