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3 years ago

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:

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