Can someone please help !! Graphing

Can someone help me figure this out?

svetlana [45]

Answer:

Il try So try to see what angle it is at

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre

vfiekz [6]

For AD:
AD=root((c-0)^2 + (d-0)^2)=root((c)^2 + (d)^2)
For BC:
BC=root(((b+c) - b)^2+(d-0)^2)=root((c)^2+(d)^2)
For AB:
AB=root((b-0)^2 + (0-0)^2)=root((b)^2 + (0)^2)=root((b)^2)
For CD:
CD=root((c-(b+c))^2 + (d-d)^2)
CD=root((b)^2 + (0)^2)
CD=root((b)^2)

8 0

3 years ago

Hey there!<br> What is rational numbers and how can I multiple them??<br> Pls tell mee!!

Elza [17]

Answer:

Just like multiplying fractions

Step-by-step explanation:

Rational numbers are like fractions a/b

and they are multiplied straight across

4 6 24 because 4 x 6 = 24

-- x -- = ------

5 7 35 because 5 x 7 = 35

8 0

3 years ago

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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

Use the construction in proof of the Chinese reminder theorem to find all solutions to the system of congruence:

antoniya [11.8K]

Answer:

17,101,185, 269,.... is the solution.

i.e. x≡17 mod(84) is the solution

Step-by-step explanation:

Given that the system is

$x ≡ 2 ( mod 3 )x ≡ 1 ( mod 4 )x ≡ 3 ( mod 7 )$

Considering from the last as 7 is big,

possible solutions would be 10,17,24,...

Since this should also be 1(mod4) we get this as 1,5,9,...17, ...

Together possible solutions would be 17, 45,73,121,....

Now consider I equation and then possible solutions are

5,8,11,14,17,20,23,26,29,...,47,....75, ....

Hence solution is 17.

Next number satisfying this would be 101, 185, ...

3 0

3 years ago