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3 years ago

Please help me?!?!!?!!??!!?!!!

Mathematics

2 answers:

juin [17]3 years ago

8 0

20+60 = 80 but then subtract 36 which = 44. Finally, add 70 and you get 114

hjlf3 years ago

4 0

I'm pretty sure it's 114 miles.

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A study found that 9% of dog owners brush their dog’s teeth. Of 578 dog owners, about how many would be expected to brush their

Amanda [17]

Answer:

A. 50

Step-by-step explanation:

Simply multiply 9% and 578 to get your answer:

578(0.09) = 52.02

5 0

3 years ago

I need help on this question

gtnhenbr [62]

8:20 pm because if Prisha read 60 pages for a half hour, and she read 40 until 8:40, it would be 8:20 pm

8 0

3 years ago

Type the correct answer in each box. Spell all words correctly, and use numerals instead of words for numbers.

Andru [333]

Answer:

Nate made a mistake. He should have got 24 as Least common factor.

Step-by-step explanation:

Given number are 8 and 12.

To find the GCF or LCF we have to list multiples.

For LCF

List the multiples of each number until at least one of the multiples appears on all lists.
Find the smallest number that is on all of the lists.
This number is the LCM.

For GCF

List the multiples of each number until at least one of the multiples appears on all lists.
Find the Biggest number that is on all of the lists.
This number is the GCF

The least common multiple of 8 and 12 is 24.

The Greatest common multiply of 8 and 12 is 4.

Hence, Nate is wrong. Nate should have found 24.

[RevyBreeze]

7 0

2 years ago

HELPP!!! <br><br><br> What is the measure of ADC

Gwar [14]

72 degrees

(well it says answers need to be 20 words long.. So I wrote this)

7 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