All categories

3 years ago

Please help I need it in 5 minutes!

Mathematics

1 answer:

lesantik [10]3 years ago

7 0

Answer:

$11^{-10}$

Step-by-step explanation:

When you divide exponents, you subtract their powers:

-5 - 5 = -10

$11^{-10}$

You might be interested in

What is 464.1587734 in fraction form

loris [4]

Answer:

464.1587734 = 464 793867/5000000

3 0

3 years ago

The general form of the equation of a circle is x^2+y^2−4x−8y−5=0.

lakkis [162]

X² - 4x + y² - 8y = 5

Complete the square:
x² - 4x + (4) + y² - 8y + (16) = 5 + (4) + (16)
→ (x - 2)² + (y - 4)² = 25

Center: (2,4)
radius: √25 = 5

5 0

3 years ago

Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

7 0

3 years ago

What is the range of the function shown on the graph above?

nikklg [1K]

Answer:

D: 0 < y < 7

Step-by-step explanation:

The domain is the input (i.e. the x values)

The range is the output (i.e. the y values)

So for the range, look at the y-coordinates of the end points of the line: (-6, 0) and (9, 7) ⇒ 0 < y < 7

4 0

2 years ago

EASY GRAPHS ILL AWARD BRAINLIEST, can someone fill it in all again for me , recordect mine if wrong

sweet-ann [11.9K]

Answer: I will answer questions answer to question question answer questions about the answer question answer questions and questions about questions answer questions questions about questions or questions about questions answer questions about the questions answer question

I have a little bit of a little bit more about to get a new one and but I’m going back to the office now and I’m not sure if you want Step-by-step explanation:

4 0

2 years ago