Mathematics, 18.08.2019 12:30 creepycrepes

Find the length of ab
a (2,-4) b (6,2)
solve for x and y
2x+3y+14
-4x+2y+4
given that b is the midpoint of segment ac and a (3,-2) and b (1,1), find c

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FIRST QUESTION
Given points are:
A(2, -4) and B(6, 2)
Now, Use the distance formula.
distance formula = $\sqrt{ (x_{2}- x_{1})^{2} + ( y_{2} - y_{1} )^{2} }$

Now, plug the values into the formula, So,
distance = $\sqrt{ (6- 2)^{2} + ( 2 - (-4))^{2} }$

= $\sqrt{ (6- 2)^{2} + ( 2 +4))^{2} }$

= $\sqrt{ (4)^{2} + ( 6))^{2} }$

= $\sqrt{ 16+36}$

= $\sqrt{52}$

= $2 \sqrt{13}$

So, the length of AB is $2 \sqrt{13}$ .

THIRD QUESTION
Two points given are:
A(3, -2) and B(1, 1)
Also given that B is the midpoint of AC.

Let, the co-ordinates of C be C(a, b).
Now, using midpoint formula,
Midpoint = $(\frac{ x_{1}+ x_{2} }{2} , \frac{ y_{1}+ y_{2} }{2} )$

$(1, 1)=(\frac{ 3+ a }{2} , \frac{ -2+b }{2} )$

Now, equaling the ordered pair, we have,

$1=\frac{ 3+ a }{2}$ .............equation (1)

$1=\frac{ -2+b }{2}$ ................equation (2)

Now, taking equation (1)
$1=\frac{ 3+ a }{2}$

1*2=3+a

Now, taking equation (2)
$1=\frac{ -2+b }{2}$

1*2=-2+b

So, the co ordinates of C are (a, b) which is (-1 , 4)

SECOND QUESTION:
Given equations are:
2x + 3y = 14.....................equation (1)
-4x + 2y = 4 .....................equation (2)
Taking equation (2)
-4x + 2y = 4
2y = 4 + 4x
y = (4 + 4x) / 2
y = 2 + 2x .......................equation (3)
Now, Taking equation (1)
2x + 3y = 14
Substituting the value of y from equation (3), we get,
2x + 3(2 + 2x) = 14
2x + 6 + 6x = 14
8x = 14 - 6
x = (14 - 6) / 8
x = 1

Taking equation (3)
y = 2 + 2x
Now, substituting the value of x in equation (3), we get,
y= 2 + 2 (1)
y = 2 + 2
y = 4

So, x=1 and y=4

Answer from: Quest

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3

Answer from: Quest

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

=3x+1

=3(2)+1

since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

=x+1

=2+1

now, let's find g(2)g(2):

g(x)

g(2)

=2x

=2⋅2

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

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Find the length of ab
a (2,-4) b (6,2)
solve for x and y
2x+3y+14
-4x+2y+4

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