Advanced Placement (AP), 24.11.2020 14:00 dreamadoreaa

Essay on your thoughts about LGBTQ COMMUNITY

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Answer from: zeesharpe05

Intro:

The LGBTQ+ comunity is a large and powerfull comunity of people who just want to be themselves without judgement. Although, whith how our world is, and with so many groups wanting this same thing, the world is full of judgement, and hate.

paragraph 1:

the title LGBTQ+ is just an acronym(an abbreviation formed from the initial letters of other words)that stands for a basic showing of some sexualitys or ways people might identify themselves. The first letter of the acronym (L) stands for Lesbian, which means someone who identifies as a woman, likes only other women. The G stands for Gay, which means someone who identifys as a male, likes only other males. now is where it gets a little more "confusing". The B stands for bisexual (or bi), which means the person likes 2 genders. you would think boy and girl, but the person who identifies as bi could also like boys and non-binary people, or girls and non-binary people, ext. The T stands for trans which means they feel like they are a other gender trapped inside a body that they were assigned at birth that is the opposite gender of who they believe they really are. The last letter in the acronym is Q which stands for queer which is just an umbrella term for someone who feels like they are a part of the LGBTQ+ community who either doesn't know what label fits them, doesn't feel comfortable using any of the terms, or just likes the of just being "queer". last but not least, the + stands for all of the other parts of the community, like pansexual, omnisexual, asexual, Polyamory, non-binary, demi boy/girl, and MANY more.

you can go off of that if you'd like :)

Answer from: Quest

You need to figure out at which x-value does the graph has a tangent line slope of -1.remember, the derivative is the slope of the tangent line. so find the derivative function and set it to equal -1 to find the x-value that will have a tangent line slope of 0.differentiate with the chain rule. $\begin{aligned} f(x) & = (0.5)^{2x} \\ f'(x) & = 0.5^{2x} \cdot \ln(0.5) \cdot (2x)' \\ f'(x) & = 0.5^{2x} \cdot \ln(0.5) \cdot 2 \\ f'(x) & = 2\ln(0.5) \cdot 0.5^{2x} \end{aligned}$ set the derivative to equal -1 and solve for x. $\begin{aligned} -1 & = 2\ln(0.5) \cdot 0.5^{2x} \\ -\frac{1}{2} & = \ln(0.5) \cdot 0.5^{2x} \\ -\frac{1}{2\ln(0.5)} & = 0.5^{2x} \end{aligned}$ take log base 10 of both sides. also recall that $\log_c(m^n) = n \log_c m$ . $\begin{aligned} \log\left[ -\frac{1}{2\ln(0.5)}\right] & = \log\left[0.5^{2x}\right] \\ \log\left[ -\frac{1}{2\ln(0.5)}\right] & = 2x \log(0.5) \\ \dfrac{\log\left[ -\frac{1}{2\ln(0.5)}\right]}{2\log(0.5)} & = x \\ x & \approx 0.2356 \end{aligned}$ so the x-coord of the point of tangency is $x \approx 0.2356$ .the y-coordinate of this point of tangency can be calculated using the original function: $\begin{aligned} f(x) & = (0.5)^{2x} \\ f(0.2356) & = (0.5)^{2 \cdot 0.2356} \\ & \approx 0.7213 \end{aligned}$ using point-slope form, the equation of the tangent line is $y - 0.7213 = -1(x-0.2356)$ the x-intercept of this tangent line is when the graph is at y=0. $\begin{aligned} 0 - 0.7213 & = -1(x-0.2356)\\ - 0.7213 & = -(x-0.2356) \\ 0.7213 & = x-0.2356\\ x & = 0.9569 \end{aligned}$ this rounds off to $x & = 0.957$ .choice c.

Answer from: Quest

answer: no

explanation:

i thought it would be hard. yeah it took a little while to get used to but in all you will be fine. just do your work and stay out of drama.

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