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Artemon [7]
3 years ago
6

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:

<u><em>4 : </em></u> Coordinate of Q = (\frac{1}{5} ,\frac{-12}{5} )

<em><u>5 : </u></em>  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 *<br><br><br>x = 1<br><br>x = 5<br><br>x = -7<br><br>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 &gt; -a, then _____.<br><br> a &gt; 2<br> a &gt; -2<br> a &lt; 2<br> a &lt; -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
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