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3 years ago

Find the missing angles

Mathematics

1 answer:

swat323 years ago

8 0

Answer:

Explained below.

Step-by-step explanation:

1. Bisecting splits an angle into two equal parts. Angle 2 and 3 are now equal.

2. And since it is a right angle, both angles are 45 degrees.

3. All triangle angles will ALWAYS add up to 180.

4. Angle 4 = 180 - 45 (Angle 3) -83.

Angle 4 = 52.

5. Do same thing for Angle 1.

6. Angle 1 = 180 - 45 (Angle 2) - 48

Angle 1 = 87.

Final answers.

1. 87 degrees

2. 45 degrees.

3. 45 degrees.

4. 52 degrees.

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Is -28 greater than or less than -162

vivado [14]

Answer:

Any number which is on left side on the line number is less than the number which is on right hand side relative to that number.

As -28 is on the right hand side of the line number, and -162 is heading towards left hand side relative to -28.

Therefore, -28 is greater than -162.

Step-by-step explanation:

To determine:

Is -28 greater than or less than -162?

Solution Steps:

In mathematics, a number line is considered to be a straight line with numbers placed at equal intervals along its length.

A number line could be extended infinitely in any direction i..e -∞ to +∞.

A number is usually represented horizontally.

Any number which is on left side on the line number is less than the number which is on right hand side relative to that number.

As -28 is on the right hand side of the line number, and -162 is heading towards left hand side relative to -28.

Therefore, -28 is greater than -162.

In other words,

$-28>-162\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:\mathrm{True\:for\:all}\:x\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}$

Keywords: number line, less than, greater than

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7 0

3 years ago

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Tomtit [17]

Answer:

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Thanks<3

Life is short, Smile while you still have teeth!!! 

8 0

2 years ago

Read 2 more answers

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 .