What is the slope of y=-x perpendicular to the equation

Mathematics

1 answer:

tensa zangetsu [6.8K]3 years ago

7 0

$\frac{1}{2} x$

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If f(x) = x-6 and g(x)= 1/2x (x+3), find g(x) * f(x)

sertanlavr [38]

Answer:

Final answer is $g\left(x\right)\cdot f\left(x\right)=\frac{\left(x-6\right)}{2x\left(x+3\right)}$ .

Step-by-step explanation:

given functions are $f(x)=x-6$ and $g\left(x\right)=\frac{1}{2x\left(x+3\right)}$ .

Now we need to find about what is the value of $g\left(x\right)*f\left(x\right)$ .

$g\left(x\right)*f\left(x\right)$ simply means we need to multiply the value of $f(x)=x-6$ and $g\left(x\right)=\frac{1}{2x\left(x+3\right)}$ .

$g\left(x\right)\cdot f\left(x\right)=\frac{1}{2x\left(x+3\right)}\cdot\left(x-6\right)$

$g\left(x\right)\cdot f\left(x\right)=\frac{\left(x-6\right)}{2x\left(x+3\right)}$

Hence final answer is $g\left(x\right)\cdot f\left(x\right)=\frac{\left(x-6\right)}{2x\left(x+3\right)}$ .

5 0

3 years ago

Arlo invested $4000 in an account that earns 5.5% interest, compounded annually. The formula for compound interest is A(t)=P(1+i

USPshnik [31]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$4000\\ r=rate\to 5.5\%\to \frac{5.5}{100}\dotfill &0.055\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases} \\\\\\ A=4000\left(1+\frac{0.055}{1}\right)^{1\cdot 4}\implies A=4000(1.055)^4\implies A\approx 4955.2986$