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3 years ago

Digits after the decimal point (with the exception of leading 0's) become the ____ of the fraction.

Mathematics

1 answer:

gayaneshka [121]3 years ago

7 0

Your answer is the numerator

The denominator would be a power of 10. Then you could simplify

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Help please!! Just need C <br><br>(don't worry about the background shapes)

horrorfan [7]

I think not fpr sure though

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3 years ago

Deluxe River Cruises operates a fleet of river vessels. The fleet has two types of vessels: A type-A vessel has 60 deluxe cabins

bixtya [17]

Answer: It should be used 2 for type-A and 3 for type-B to minimize the cost.

Step-by-step explanation: As it is stipulated, <u>x</u> relates to type-A and y to type-B.

Type-A has 60 deluxe cabins and B has 80. It is needed a minimum of 360 deluxe cabins, so:

60x + 80y ≤ 360

For the standard cabin, there are in A 160 and in B 120. The need is for 680, so:

160x + 120y ≤ 680

To calculate how many of each type you need:

60x + 80y ≤ 360

160x + 120y ≤ 680

Isolating x from the first equation:

x = $\frac{360 - 80y}{60}$

Substituing x into the second equation:

160( $\frac{360 - 80y}{60}$ ) + 120y = 680

-3200y+1800y = 10200 - 14400

1400y = 4200

y = 3

With y, find x:

x = $\frac{360 - 80y}{60}$

x = $\frac{360 - 80.3}{60}$

x = 2

To determine the cost:

cost = 42,000x + 51,000y

cost = 42000.2 + 51000.3

cost = 161400

To keep it in a minimun cost, it is needed 2 vessels of Type-A and 3 vessels of Type-B, to a cost of $161400

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3 years ago

A home improvement store advertises 50 square feet of flooring for $300.00 including a $100.00 installation fee. How much does e

dimulka [17.4K]

$6 . It will cost $6 because spending 6 dollars , 50 times will equal $300

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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 .

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3 years ago

Please help!!!! Will mark Brainliest!!!! Pleaseeee!! 20 Points!!!!

Lesechka [4]

Your answer : is b (the letter b)

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3 years ago