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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

What is the perimeter of a polygon with verticals at (-1, 3) , (-1, 6) , (2, 10) , (5, 6) , and (5, 3)

Marat540 [252]

Answer:

22 units

Step-by-step explanation:

The perimeter of a polygon is said to be the sum of the length of it's sides.

From the question, we have 5 vertices. This means the polygon is a pentagon. It's given vertices are

A = (−1, 3)

B = (−1, 6)

C = (2, 10)

D = (5, 6)

E = (5, 3)

To find the distance between two points, we use the formula

d = √[(y2 - y1)² + (x2 - x1)²]

Between A and B, we have

d(ab) = √[(6 - 3)² + (-1 --1)²]

d(ab) = √(3²) + 0

d(ab) = √9 = 3

Between B and C, we have

d(bc) = √[(10 - 6)² + (2 --1)²]

d(bc) = √[4² + 3²]

d(bc) = √(16 + 9) = √25 = 5

Between C and D, we have

d(cd) = √[(6 - 10)² + (5 - 2)²]

d(cd) = √[(-4)² + 3²]

d(cd) = √(16 + 9) = √25 = 5

Between D and E, we have

d(de) = √[(3 - 6)² + (5 - 5)²]

d(de) = √(-3)² + 0

d(de) = √9 = 3

Between E and A, we have

d(ea) = √[(3 - 3)² + (5 --1)²]

d(ea) = √[0 + (6)²]

d(ea) = √36 = 6

The perimeter is given as

d(ab) + d(bc) + d(cd) + d(de) + d(ea) =

3 + 5 + 5 + 3 + 6 = 22 units

7 0

2 years ago

Help please ASAP thank you

jonny [76]

Question 1

If we let $FE=x$ , then $DE=x-5$ .

Also, as $\overline{DF}$ bisects $\overline{BC}$ , this means $BE=EC=6$ .

Thus, by the intersecting chords theorem,

$6(6)=x(x-5)\\\\36=x^2 - 5x\\\\x^2 - 5x-36=0\\\\(x-9)(x+4)=0\\\\x=-4, 9$

However, as distance must be positive, we only consider the positive case, meaning FE=9

Question 2

If we let CE=x, then because AB bisects CD, CE=ED=x.

We also know that since FB=17, the radius of the circle is 17. So, this means that the diameter is 34, and as AE=2, thus means EB=32.

By the intersecting chords theorem,

$2(32)=x^2\\\\64=x^2\\\\x=-8, 8$

However, as distance must be positive, we only consider the positive case, meaning CE=8

4 0

2 years ago

Help plz I need big help someone plz help me with this question!

ozzi

Answer:

Step-by-step explanation to be honest probably 12 if not i’m so sorry

3 0

2 years ago

A function is given. Determine the average rate of change of the function between the given values of the variable: f(x)= 4x^2;

nevsk [136]

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------$

$\bf f(x)= 4x^2 \qquad 
\begin{cases}
x_1=3\\
x_2=3+h
\end{cases}\implies \cfrac{f(3+h)-f(3)}{(3+h)~-~(3)}
\\\\\\
\cfrac{[4(3+h)^2]~~-~~[4(3)^2]}{\underline{3}+h-\underline{3}}\implies \cfrac{4(3^2+6h+h^2)~~-~~4(9)}{h}
\\\\\\
\cfrac{4(9+6h+h^2)~~-~~36}{h}\implies \cfrac{\underline{36}+24h+4h^2~~\underline{-~~36}}{h}
\\\\\\
\cfrac{24h+4h^2}{h}\implies \cfrac{\underline{h}(24+4h)}{\underline{h}}\implies 24+4h$

8 0

3 years ago