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3 years ago

The side length of a square sticker is 3 inches. How many of these

Mathematics

1 answer:

DENIUS [597]3 years ago

5 0

Answer:

just divide 3 and 16

Step-by-step explanation:

u will get ur answer then

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A person who is 5 feet tall is standing 88 feet from the base of a tree, and the tree casts a 96 foot shadow. the person's sha

pashok25 [27]

5/8=x/96 96÷8=12 12×5=60

3 0

4 years ago

100 POINTS ANSWER FAST NO BEGINNERS OR AMBITIOUS, VIRTUOSO AND UP ONLY

Flauer [41]

I hope this helps you

8 0

3 years ago

Read 2 more answers

What value of Y will satisfy the given equation y divided by 2y — 15 = 7/5

guapka [62]

Answer:

$\huge\boxed{\sf y = \frac{41}{5} \ OR \ y = 8\frac{1}{5} }$

Step-by-step explanation:

$\sf 2y-15 = \frac{7}{5} \\\\Adding \ 15 \ to \ both \ sides\\\\2y = \frac{7}{5} + 15\\\\2y = \frac{7+75}{5} \\\\2y = \frac{82}{5} \\\\Dividing \ both \ sides \ bu 2\\\\y = \frac{82}{5*2} \\\\y = \frac{41}{5}$

Hope this helped!

<h2>~AnonymousHelper1807</h2>

3 0

3 years ago

Is this answer right?

zalisa [80]

Answer:

yes I think so

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

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