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2 years ago

Which side lengths form a right triangle?

Mathematics

2 answers:

saw5 [17]2 years ago

8 0

B
because
${1}^{2} + {8}^{2} = {( \sqrt{65}) }^{2}$

ludmilkaskok [199]2 years ago

5 0

If you get a true statement when you simplify, then you do indeed have a right triangle! If you get a false statement, then you can be sure that your triangle is not a right triangle. Check out this tutorial and learn how use the Pythagorean theorem to see if a triangle is a right triangle!

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What is 2 X (6+ 2 + 8) -4

VladimirAG [237]

Answer:

The correct answer is 32x - 4

Step-by-step explanation:

1) Simplify 6 + 2 to 8.

$2x(8 + 8) - 4$

2) Simplify 8 + 8 to 16.

$2x \times 16 - 4$

3) Simplify 2x × 16 to 32x.

$32x - 4$

Therefor, the answer is 32x - 4.

4 0

2 years ago

the cost of belonging to a gym can be modeled by by c(m)=50m+79.50 what is the meaning of the slope and y-intercept

Bess [88]

The slope is the relationship between the cost of the gym and how long you are there for.
the y-intercept is how much it costs to go to the gym for no hours ie the start-up cost.

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

2 years ago

You are an accountant doing an audit for a law firm. A paralegal worked for a client for 5 hours per day for 3 days. The client

blagie [28]

Total hours worked = 5 x 3 = 15 hours

Amount collected = 50 x 15 = $750
Amount paid to the paralegal = 30 x 15 = $450
Amount made by the firm = 750 - 450 = $300

3 0

3 years ago

Read 2 more answers

!!!*BRAINLIEST*!!! Solve for the value of e

Elena-2011 [213]

It is really easy it is 6

4 0

2 years ago

Read 2 more answers