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cluponka [151]
3 years ago
14

Please help !!!! find the tangent of angle (-)

Mathematics
1 answer:
Troyanec [42]3 years ago
8 0
15/8. Tangent = opposite / adjacent line
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How many ways can Marie choose 2 pizza toppings from a menu of 6 toppings if each topping can only be chosen once?
m_a_m_a [10]

Answer:

30 different ways

Step-by-step explanation:

Hi

for the first topping she can choose 1 of 6 items

so she has 6 choices

and then the second topping she has 5 choices as each topping can only be chosen once

so it comes 6 * 5 = 30

5 0
3 years ago
Four bread rolls weigh a total of 3/5 pounds. houw much does each bread roll weigh?​
NikAS [45]

Answer:

each bread roll weighs 3/20 pounds. (you can also write it as 0.15 pounds)

Step-by-step explanation:

4 0
3 years ago
5x+y=27 y=4x using substitution method
Dvinal [7]
 5x + y = 27
substitute: 5x + 4x = 27
combine: 9x = 27
divide (by 9): x = 3
4 0
3 years ago
Read 2 more answers
Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"
vampirchik [111]
\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h

Employ a standard trick used in proving the chain rule:

\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h

The limit of a product is the product of limits, i.e. we can write

\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)

The rightmost limit is an exercise in differentiating \sqrt x using the definition, which you probably already know is \dfrac1{2\sqrt x}.

For the leftmost limit, we make a substitution y=\sqrt x. Now, if we make a slight change to x by adding a small number h, this propagates a similar small change in y that we'll call h', so that we can set y+h'=\sqrt{x+h}. Then as h\to0, we see that it's also the case that h'\to0 (since we fix y=\sqrt x). So we can write the remaining limit as

\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}

which in turn is the derivative of \tan y, another limit you probably already know how to compute. We'd end up with \sec^2y, or \sec^2\sqrt x.

So we find that

\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}
7 0
3 years ago
What is 3ten thousand+14ten+16one
ozzi
3,000 + 14 + 16 3,030
5 0
3 years ago
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