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2 years ago

The Rodriguez

Mathematics

1 answer:

Soloha48 [4]2 years ago

5 0

1,875 - 895 = 980 dollars they spent. 980 divided by 140 equals 7. They have been on their vacation for 7 days.

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if the rate of change for one linear function is positive and for another is negative can they both be either increasing or decr

Alla [95]

Answer:

They can never be both either increasing or decreasing.

Step-by-step explanation:

If the rate of change i.e. the slope of one linear function is positive, that means the graph of the linear function makes angle which varies between 0° to 90° with respect to the positive direction of the x-axis.

Therefore, the function must be increasing.

Again, if the rate of change i.e. the slope of one linear function is negative, that means the graph of the linear function makes angle which varies between 90° to 180° with respect to the positive direction of the x-axis.

Therefore, the function must be decreasing.

Hence, if the rate of change of one linear function is positive and for another is negative, they can never be both either increasing or decreasing. (Answer)

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3 years ago

You owe your mom 10$. You pay her back 7$. Write an expression that can be used to represent this situation.

mezya [45]

Answer:

10$-7$=3$

Step-by-step explanation:

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3 years ago

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HELP QUICK! NO BOTS!

gtnhenbr [62]

Answer:I'm pretty sure it's A

Step-by-step explanation:

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2 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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2 years ago

ANSWER NUMBER 11 PLZ AND ILL MARK YOU BRAINLIEST !!!

Yakvenalex [24]

Answer:

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3 years ago

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