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2 years ago

Question within image.

Mathematics

1 answer:

julia-pushkina [17]2 years ago

6 0

1)a parallelogram is a quadrilateral with opposite sides being parallel so the lines AB must equal CD or otherwise it wouldn't be a parallelogram

2)for a quadrilateral to be proven to be a parallelogram the consecutive angles has to be supplementary which means they add up to 180 or for example you could slice the parallelogram in half down the middle you could piece the two consecutive angles together to make 180 degrees

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Can someone please help me with number 2 please I’ll love you forever ;)

yawa3891 [41]

Step-by-step explanation:

the first question

x²+x+5

when x is -4

(-4)²+(-4)+5

16-4+5

12+5

hope this was helpful

7 0

2 years ago

Let's Create

Cerrena [4.2K]

Answer:

1) 1/4+___=7/12

2) 5/8-___= -3/8

3) 1/3÷___= 1/9

4) (2.1)(____)=0.63

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3 0

2 years ago

1) find the equation of the line parallel to

In-s [12.5K]

Answer:

1) The equation of the line parallel to x-5y=6 and through (4,-2) is 5y = x -14

2) The equation of the line perpendicular to y= -2/5x + 3 and through (2,-1) is 2y = 5x -12

Solution:

1) find the equation of the line parallel to x-5y=6 and through (4,-2).

Given, line equation is x – 5y = 6

We have to find the line equation that is parallel to given line and passing through the point (4, -2)

Now, let us find slope of the given line.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-5}=\frac{1}{5}$

Now, we know that, slope of parallel lines are equal.

So, slope of required line is 1/5 and it passes through (4, -2)

Now, using point slope form

$y-y_{1}=m\left(x-x_{1}\right) \text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on line. }$

$y-(-2)=\frac{1}{5}(x-4) \rightarrow 5(y+2)=x-4 \rightarrow 5 y+10=x-4 \rightarrow x-5 y=14$

Hence, the line equation is 5y = x -14

2) find the equation of the line perpendicular to y= -2/5x + 3 and through (2,-1)

$\text { Given, line equation is } y=-\frac{2}{5} x+3 \rightarrow 5 y=-2 x+15 \rightarrow 2 x+5 y=15$

We have to find the line equation that is perpendicular to given line and passing through the point (2, -1)

Now, let us find slope of the given line.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-2}{5}=-\frac{2}{5}$

Now, we know that, product of slopes of perpendicular lines equals to -1.

So, slope of required line $\times$ slope of given line = -1

slope of required line = $-1 \times \frac{5}{-2}=\frac{5}{2}$

And it passes through (2, -1)

Now, using point slope form

$\begin{array}{l}{\text { Line equation is } y-(-1)=\frac{5}{2}(x-2) \rightarrow 2(y+1)=5(x-2)} \\\\ {\rightarrow 2 y+2=5 x-10 \rightarrow 5 x-2 y=12}\end{array}$