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Mathematics

2 answers:

zlopas [31]3 years ago

7 0

The answer is a BDEFG

Artist 52 [7]3 years ago

3 0

The answer is going to be BDEFG

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Find the Y-coordinate of point P that lies 1/3 along segment RS, where R (-7, -2) and S (2, 4).

QveST [7]

Solution:

Given that the point P lies 1/3 along the segment RS as shown below:

To find the y coordinate of the point P, since the point P lies on 1/3 along the segment RS, we have

$\begin{gathered} RP:PS \\ \Rightarrow\frac{1}{3}:\frac{2}{3} \\ thus,\text{ we have} \\ 1:2 \end{gathered}$

Using the section formula expressed as

$[\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}]$

In this case,

$\begin{gathered} m=1 \\ n=2 \end{gathered}$

where

$\begin{gathered} x_1=-7 \\ y_1=-2 \\ x_2=2 \\ y_2=4 \end{gathered}$

Thus, by substitution, we have

$\begin{gathered} [\frac{1(2)+2(-7)}{1+2},\frac{1(4)+2(-2)}{1+2}] \\ \Rightarrow[\frac{2-14}{3},\frac{4-4}{3}] \\ =[-4,\text{ 0\rbrack} \end{gathered}$

Hence, the y-coordinate of the point P is

$0$

8 0

1 year ago

HELP ME Using the following equation, complete the square and find the center and radius of the circle:

IrinaK [193]

Answer: $(-2,3)$ , $r=5\ \text{units}$

Step-by-step explanation:

Given

the equation is $x^2+y^2+4x-6y-12=0$

completing the square

$\Rightarrow (x^2+2\times 2\times x+4-4)+(y^2-2\times3\times y+9-9)-12=0\\\Rightarrow (x^2+2\times 2\times x+4)+(y^2-2\times3\times y+9)-12-4-9=0\\\Rightarrow (x+2)^2+(y-3)^2=25\\\Rightarrow (x+2)^2+(y-3)^2=5^2\\\Rightarrow (x-(-2))^2+(y-3)^2=5^2$