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3 years ago

Please help ! I don’t understand :(

Mathematics

1 answer:

Elina [12.6K]3 years ago

7 0

Answer:

A. The number quadrupled each week

Step-by-step explanation:

f(x)= 3(4)^x

As per the function, each week number increases 4 times

So the answer choice is A

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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

3 years ago

Can somebody please help me here

andrey2020 [161]

The ordered pairs already plotted have x-coordinates: -3, -2, 1, and 2.

A function cannot have two ordered pairs with the same x-coordinate.

The answer cannot have -3, -2, 1, or 2 as the x-coordinate.

Answer: First choice (-1, 2)

6 0

3 years ago

Read 2 more answers

Karan needs to choose between two gym plans. He can either pay a $150 joining fee and a $10 monthly fee, or he can pay a $50 joi

Ierofanga [76]

Let x be the month on which the fee for two plans are equal.

Therefore,

150 + 10x = 50 + 30x

10x - 30x = 50 - 150

-20x = -100

x = 5

Cumulative cost = 150 + 10(5)

= 150 + 50

= 200

Hence, Karan will pay a cumulative cost of $200 dollars with either plan to go to the gym for 5 months.

4 0

3 years ago

Please help with part A!!

zaharov [31]

Answer:

Option A. A and D

Step-by-step explanation:

This answer because they are both right triangles and are near the same size

Hope this helps!

7 0

3 years ago

Solve the initial value problem. dy/dt = 1 + 6/t , t > 0, y = 8 when t = 1

nordsb [41]

$\displaystyle\frac{dy}{dt} = 1 + \frac{6}{t}\ \Rightarrow\ dy = \left( 1 + \frac{6}{t}\right) dt\ \Rightarrow\int 1\, dy = \int \left( 1 + \frac{6}{t}\right) dt \ \Rightarrow \\ \\
y = t + 6\ln|t| + C. \text{ But $t\ \textgreater \ 0$ so }y = t + 6\ln t + C. \\ \\ 
y(1) = 8\ \Rightarrow\ 8 = 1 + 6 \ln 1 + C \ \Rightarrow\ C = 7 \text{ so } \\ \\
y(t) = t + 6\ln t + 7$

5 0

3 years ago