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3 years ago

Problems this is so hard so help

Mathematics

1 answer:

lozanna [386]3 years ago

6 0

Giant spider crab are further from the surface the wolf because -800 is farther down than -300 let say the surface is 0 going down is negative you react -300 before -800

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in the year 2010, the luxury bike industry had two bike manufactures splendor and passion with the market shares of 30% and 70%,

kari74 [83]

Answer: 50%

Step-by-step explanation:

In 2010, there were only 2 bike manufacturers and they had market shares of 30% and 70%.

A new manufacturer joins the market and captures 10% of the market share and Splendor increases to 40% of market share.

Market shares:

Yamaha = 10%

Splendor = 40%

Passion = 100% - 10% - 40%

= 50%

8 0

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 .

5 0

3 years ago

Answer these below. its for algerbra math

pantera1 [17]

Answer:

lin

Step-by-step explanation:

3 0

3 years ago

If the dimensions of a square based pyramid are tripled what happens to the volume?

11Alexandr11 [23.1K]

Plug in numbers. if originally, it was V=1/2 * 1 * 1 * 2=1

<span>tripling the dimensions would be V=1/2 * 3 * 3 * 6=27 </span>

<span>it would be 27 times larger. this could also be found by 3^3, just as doubling would render 2^3

</span>

7 0

3 years ago

Read 2 more answers

If the terminal side of angle θ passes through a point on the unit circle in the first quadrant where x=√2/2, what is the exact

Serhud [2]

The exact measure of the angle is 45°.

<h3>How to get the angle?</h3>

We know that the terminal side passes through a point of the form (√2/2, y).

Notice that the point is on the unit circle, so its module must be equal to 1, so we can write:

$1 = \sqrt{( \frac{\sqrt{2} }{2} )^2 + y^2} \\\\1^2 = \frac{2}{4} + y^2\\1 - 1/2 = y^2\\\\1/\sqrt{2} = y$

We know that y is positive because the point is on the first quadrant.

Now, we know that our point is:

(√2/2, 1/√2)

And we can rewrite:

√2/2 = 1/√2

So the point is:

( 1/√2, 1/√2)

Finally, remember that a point (x, y), the angle that represents it is given by:

θ = Atan(y/x).

Then in this case, we have:

θ = Atan(1/√2/1/√2) = Atan(1) = 45°

If you want to learn more about angles, you can read:

brainly.com/question/17972372

3 0

2 years ago