If T = 10nz is solved for z then z =?

Mathematics

1 answer:

Vlad [161]3 years ago

8 0

T = 10nz

Hope this will help...

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12) Given the t-chart below,determine if the number

Goryan [66]

Hi! I'm going to assume that your chart went something like 1 ride is 20 dollars, 2 rides are 60, and 3 rides are 40 dollars.

Proportional relationships are when the proportions between the two variables x and y are constant. Thinking of a chocolate being the same price per bar no matter how much you buy of it in bulk might help you out.

In this case, one ride is 20 dollars. However, if I were to buy two rides, it would be 30 dollars per ride. In the last ride, rides are 13.33 dollars for 3 rides. The price of each ride for each variable is different, therefore this would not be a proportional relationship.

6 0

2 years ago

I need help with this question! Quickly please!

sleet_krkn [62]

Answer:

The answer is C. ASA

Step-by-step explanation:

The SAS doesn't work here, and there are not two angles right next to each other to use.

8 0

3 years ago

Read 2 more answers

Use an Addition or Subtraction Formula to find the exact value of the expression, as demonstrated in Example 1. cos 41π 12.

PtichkaEL [24]

Answer: $\frac{\sqrt{6}-\sqrt{2}}{2}$

Step-by-step explanation:

We apply the formula $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$ .

Note that $\cos(\frac{41}{12}\pi)=\cos((\frac{36}{12}+\frac{7}{12})\pi)=\cos(3\pi + \frac{7}{12})\pi)$ . Take $x=3\pi$ and $y=\frac{7}{12}\pi$ in the formula above to get

$\cos(\frac{41}{12}\pi)=\cos(3\pi)\cos(\frac{7}{12}\pi)-\sin(3\pi)\sin(\frac{7}{12}\pi)=(-1)\cdot \cos(\frac{7}{12}\pi)-0\cdot\sin(\frac{7}{12}\pi)=-\cos(\frac{7}{12}\pi)$

Then the value of this expression is $-\cos(\frac{7}{12}\pi)$

We can use the cosine addition formula again to simplify further. Decompose the fraction in the argument as:

$\cos(\frac{7}{12}\pi)=\cos((\frac{3}{12}+\frac{4}{12})\pi)=\cos((\frac{1}{4}\pi + \frac{1}{3})\pi)$

Applying the formula with $x=\frac{1}{4}\pi$ and $y=\frac{1}{3}\pi$ we obtain

$\cos(\frac{7}{12}\pi)=\cos(\frac{1}{4}\pi)\cos(\frac{1}{3}\pi)-\sin(\frac{1}{4}\pi)\sin(\frac{1}{3}\pi)=\frac{\sqrt{2}}{2}\cdot\frac{1}{2} -\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{2}-\sqrt{6}}{2}$