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3 years ago

Anyone trying to run some fortnite

Mathematics

2 answers:

ioda3 years ago

5 0

Answer:

I can probably play later

Step-by-step explanation:

Im a lvl 44 almost 45 though, add me Auroras Lights

Ann [662]3 years ago

3 0

Answer:

I already ran over fortnite

Step-by-step explanation:

You might be interested in

Suppose that 40 deer are introduced in a protected wilderness area. The population of the herd

Scorpion4ik [409]

The horizontal asymptote of the function is the minimum number of deer in the area.

The equation of horizontal asymptote is: $\mathbf{y = 40}$
The horizontal asymptote means that, the number of deer will never be less than 40

The equation is given as:

$\mathbf{f(x) = \frac{40 + 2x}{1 + 0.05x}}$

Expand the numerator

$\mathbf{f(x) = \frac{40(1 + 0.05x)}{1 + 0.05x}}$

Cancel out the common factor

$\mathbf{f(x) = 40}$

Hence, the equation of horizontal asymptote is:

$\mathbf{y = 40}$

The horizontal asymptote means that, the number of deer will never be less than 40

Read more about horizontal asymptotes at:

brainly.com/question/4084552

8 0

3 years ago

Choose the table that represents g(x) = 3⋅f(x) when f(x) = x − 1.

Nastasia [14]

$\bf f(x)=x-1\qquad \qquad g(x)=3\cdot f(x)\implies g(x)=\underline{3(x-1)} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{ccll} x&g(x)\\ \cline{1-2} 1&0\\2&3\\3&6 \end{array}\qquad \qquad (\stackrel{x_1}{1}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{6})$

$\bf slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-0}{3-1}\implies \cfrac{6}{2}\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-0=3(x-1)\implies y=\underline{3(x-1)}$

7 0

3 years ago

Read 2 more answers

Use the Newton-Raphson method to find the root of the equation f(x) = In(3x) + 5x2, using an initial guess of x = 0.5 and a stop

xxMikexx [17]

Answer with explanation:

The equation which we have to solve by Newton-Raphson Method is,

f(x)=log (3 x) +5 x²

$f'(x)=\frac{1}{3x}+10 x$

Initial Guess =0.5

Formula to find Iteration by Newton-Raphson method

$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\\\\x_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0})}\\\\ x_{1}=0.5-\frac{\log(1.5)+1.25}{\frac{1}{1.5}+10 \times 0.5}\\\\x_{1}=0.5- \frac{0.1760+1.25}{0.67+5}\\\\x_{1}=0.5-\frac{1.426}{5.67}\\\\x_{1}=0.5-0.25149\\\\x_{1}=0.248$

$x_{2}=0.248-\frac{\log(0.744)+0.30752}{\frac{1}{0.744}+10 \times 0.248}\\\\x_{2}=0.248- \frac{-0.128+0.30752}{1.35+2.48}\\\\x_{2}=0.248-\frac{0.17952}{3.83}\\\\x_{2}=0.248-0.0468\\\\x_{2}=0.2012$

$x_{3}=0.2012-\frac{\log(0.6036)+0.2024072}{\frac{1}{0.6036}+10 \times 0.2012}\\\\x_{3}=0.2012- \frac{-0.2192+0.2025}{1.6567+2.012}\\\\x_{3}=0.2012-\frac{-0.0167}{3.6687}\\\\x_{3}=0.2012+0.0045\\\\x_{3}=0.2057$

$x_{4}=0.2057-\frac{\log(0.6171)+0.21156}{\frac{1}{0.6171}+10 \times 0.2057}\\\\x_{4}=0.2057- \frac{-0.2096+0.21156}{1.6204+2.057}\\\\x_{4}=0.2057-\frac{0.0019}{3.6774}\\\\x_{4}=0.2057-0.0005\\\\x_{4}=0.2052$