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2 years ago

(30 points! I just need someone to correctly answer this, please.)

Mathematics

1 answer:

Art [367]2 years ago

7 0

Step-by-step explanation:

y = ⅓x − 2

y = -3x − 12

The first line has a y-intercept of -2 and a slope of ⅓.

The second line has a y-intercept of -12 and a slope of -3.

The graph looks like this: desmos.com/calculator/raouxrikbg

From the graph, we see they intersect at (-3, -3).

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Jeremy has $5.30 in nickels.

Delvig [45]

Answer: I think the answer is b

Step-by-step explanation:

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3 years ago

Read 2 more answers

Determine which is greater, the length of a square whose area is 36 in squared or the height of a cube whose volume is 125 in cu

blagie [28]

I dont really the answer i am sorry i am hust writing this response to get points and ask questions

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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

2 years ago

Pls help i will give brainliest

Novosadov [1.4K]

<h3>Answer: -a > -b</h3>

Explanation:

Let's go over an example.

We'll have a = 1 and b = 7 which makes a < b true. Feel free to pick your own favorite pair of numbers such that b is larger than 'a'.

The corresponding opposites are: -a = -1 and -b = -7

The relationship flips because the opposites of each number are flipped. So instead of saying -a < -b, we have -a > -b

Sure enough, -1 > -7 is a true statement. Use a number line to see this. I recommend a vertical number line.

In short, going from a < b to -a > -b means we flip the signs of each term, and we flip the inequality sign as well.

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2 years ago

Giving Brainliest!! 1.Zeus Industries bought a computer for $2857. It is expected to depreciate at a rate of 24% per year. What

iVinArrow [24]

2857*.24%=685.68

685.68*3=2057.04

2857-2057.04=799.96

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3 years ago