Umm anyone again lol ?

Mathematics

2 answers:

vodka [1.7K]2 years ago

7 0

I think its C.2/15 if its wrong I'm sorry.

natta225 [31]2 years ago

5 0

I'm pretty sure the answer is c, hope this helps!

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The terminal side of an angle in standard position passes through p(–3, –4). what is the value of tangent theta?

astraxan [27]

Answer:

4/3

Step-by-step explanation:

The tangent of any angle (θ) in standard position that has point (x, y) on its terminal ray is ...

tan(θ) = y/x

For the given point on the terminal side, the tangent is ...

tan(θ) = (-4)/(-3) = 4/3

_____

<em>Additional comment</em>

There are several ways this can be explained. One of them makes use of the relation between rectangular and polar coordinates:

(x, y) = (r·cos(θ), r·sin(θ))

Then the ratio y/x is ...

y/x = (r·sin(θ))/(r·cos(θ)) = sin(θ)/cos(θ) = tan(θ)

6 0

1 year ago

Read 2 more answers

Pls help it’s easy but I been staring at this for 2 hours now pls help a lot of points and will give brainlyest

diamong [38]

Answer:?

I have no idea sir sorry yt ;-;

Step-by-step explanation:

5 0

2 years ago

Find all solutions to each quadratic equation

romanna [79]

I hope this helps you

2x^2-4x+7=0 a=2 b=-4 c= 7

disctirminant =b^2-4ac

disctirminant =(-4)^2-4.2.7

disctirminant = 16-56= -40

x=-b+square root of disctirminant ÷2a

x=4+2square root of (-10)/4

x=2+square root of (-10)/2

x'=4 -2 square root of (-10)/4

x'=2 -square root of (-10)/2

6 0

2 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.