Easy Math, please help!

Mathematics

1 answer:

Lelechka [254]2 years ago

8 0

Answer:

yes

Step-by-step explanation:

angles abc and angles rqp are corresponding because they are both congruent because they both equal to 180 degrees.

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A straight line, L, is perpendicular to the straight line y=2x-1 and passes through the point (8,1). Find an equation of L

luda_lava [24]

Let's call that other line with the eqn. y=2x-1 line M.
This line is in slope-intercept form, which means that it is written in the form y=mx+b where m is the slope and b is the y-intercept.
This means that the slope of line M is 2.

Perpendicular lines have slopes which are opposite reciprocals.
(that is to say, if you flipped the fraction and changed the sign)

Of course, 2 isn't a fraction, but it's implied as 2/1.
The opposite reciprocal would then be -1/2.
Let's plug this into our slope-intercept form equation for line L.

y = mx + b
y = -1/2x + b

Of course, we still need to find that y-intercept. (y when x = 0)
To do this, we need to interpret the slope.
Slopes are rise over run, so m = -1/2 means a change of -1 in y = 2 in x.
Let's take a point we know is in our line, (8, 1).
To find that y-intercept, we want x to be 0.
To do this, we'd have to subtract 8 from x.
And according to our slope, this means adding 4 to y.
Our y-intercept is at (0, 5), with the value b that we use being just 5.

$\boxed{y=-\frac{1}2x+5}$

7 0

3 years ago

Prove that H c G is a normal subgroup if and only if every left coset is a right coset, i.e., aH = Ha for all a e G

Kaylis [27]

$\Rightarrow$

Suppose first that $H\subset G$ is a normal subgroup. Then by definition we must have for all $a\in H$ , $xax^{-1} \in H$ for every $x\in G$ . Let $a\in G$ and choose $(ab)\in aH$ ( $b\in H$ ). By hypothesis we have $aba^{-1} =abbb^{-1}a^{-1}=(ab)b(ab)^{-1} \in H$ , i.e. $aba^{-1}=c$ for some $c\in H$ , thus $ab=ca \in Ha$ . So we have $aH\subset Ha$ . You can prove $Ha\subset aH$ in the same way.

$\Leftarrow$

Suppose $aH=Ha$ for all $a\in G$ . Let $h\in H$ , we have to prove $aha^{-1} \in H$ for every $a\in G$ . So, let $a\in G$ . We have that $ha^{-1} =a^{-1}h'$ for some $h'\in H$ (by the hypothesis). hence we have $aha^{-1}=h' \in H$ . Because $a$ was chosen arbitrarily we have the desired .