Solve the missing tables part 2.

Help! It says that I need to determine If both equations represent lines which are parallel, perpendicular, or none...

12345 [234]

4x + 2y = 8 (1)
 8x + 4y = -4y (2)

A) Two lines are parallel if they have the same gradient
- putting both equations into the gradient- intercept form ( y = mx + c where m is the gradient)
(1) 4x + 2y = 8
 2y = 8 - 4x
y = -2x + 4

 (2) 8x + 4y = -4y
 8x = -4y - 4y
y = $\frac{-8x}{-8}$
y = -x
Thus the gradient of the two equations are different and as such the two lines are not parallel

B) When two lines are perpendicular, the product of their gradient is -1
$m_{1} * m_{2} = p$
p = (-2) * (-1)
p = 2
 ∴ the two lines are not perpendicular either.

Thus these lines are SKEWED LINES

4 0

2 years ago

Find the values ofx and y for the triange. x 60° 30° у

egoroff_w [7]

Answer:

x = 42, y = 36.37

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Consider the function f(x)= 2/5x-4

Gre4nikov [31]

Answer: a) $\frac{5}{2}x+10=f^{-1}(x)=g(x)$

Step-by-step explanation:

Since we have given that

$f(x)=\frac{2}{5}x-4$

a.) Find the inverse of f(x) and name it g(x).

Let f(x) = y

So, it becomes

$y=\frac{2}{5}x-4$

Switching x to y , we get

$x=\frac{2}{5}y-4$

$5x=2y-20\\\\5x+20=2y\\\\\frac{5x+20}{2}=y\\\\\frac{5}{2}x+10=y\\\\\frac{5}{2}x+10=f^{-1}(x)=g(x)$

b) . Use composition to show that f(x) and g(x) are inverses of each other.

$\mathrm{For}\:f=\frac{2}{5}x-4\:\\\\\mathrm{substitute}\:x\:\mathrm{with}\:g\left(x\right)=\frac{5}{2}x+10\\\\=\frac{2}{5}\left(\frac{5}{2}x+10\right)-4\\\\=x$

Similarly,

$\mathrm{g\left(x\right)=\frac{5}{2}x+10,\:f\left(x\right)=\frac{2}{5}x-4,\:g\left(x\right)\circ \:f\left(x\right)}\\\\\mathrm{For}\:g=\frac{5}{2}x+10\:\mathrm{substitute}\:x\:\mathrm{with}\:f\left(x\right)=\frac{2}{5}x-4\\\\=\frac{5}{2}\left(\frac{2}{5}x-4\right)+10\\\\=x$