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mariarad [96]
2 years ago
8

Here at Karen University we offer the finest education! From language lessons, now repeat after me “I want to speak to the manag

er!” “____” very good! Biology lessons, now this is your stereotypical manager! And even fashion lessons now you want the back to be as short as possible! Enroll today to become the the best pain in the asś you can be
(haha sorry)
Mathematics
1 answer:
GaryK [48]2 years ago
3 0

Answer:okkk

Step-by-step explanation:

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A board game that normally costs $30 is on sale for 25 percent off. What is the sale price of the game?
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find the discount

$30 x 0.25 = $7.50 discount of 25% off

subtract the discount  $30 - $7.50 = $22.50


The sale price of the game is $22.50.

4 0
3 years ago
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A triangular pane of glass has a height of 30 inches and an area of 270 square inches. what is the length of the base pane
Tems11 [23]
The equation for the area of a triangle is A= b*h /2. So, plug in your known variables, and you get 270in^2= b*30 /2. First you multiply by 2 to cancel out the division for 2, which gets you 540= b* 30. To get "b" by itself, divide both sides by 30, giving you <u>18=b</u>.

To check this, plug in your new known variable, as 270= 18* 30 /2. This will equal as 270=270, confirming 18in as the correct measurement.
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2 years ago
a section of a stained glass window is shaped like a parallelogram. Its base is 6.5 inches, and its height is 4 inches. how much
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8 0
3 years ago
Which of the following is a fully factored expression of 24x – 56y + 72?
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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 .

 

5 0
2 years ago
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