2 years ago

Let (h) be the interse of f (x)= 4x+ 1, which of the following representes f (x)?

Mathematics

1 answer:

liraira [26]2 years ago

5 0

Answer:

» Answer: The third objective

${ \tt{h(x) = - 4x + 1}}$

• make x the subject:

${ \tt{h(x) - 1 = - 4x }} \\ \\ { \tt{x = \frac{h(x)}{ - 4} - \frac{1}{ - 4} }} \\ \\ { \tt{x = - \frac{h(x)}{4} + \frac{1}{4} }}$

• Therefore, f(x):

${ \boxed{ \boxed{ \tt{ \: \: f(x) = - \frac{1}{4}x + \frac{1}{4} \: \: }}}}$

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Suppose a and b are both non zero real numbers. Find real numbers c and d such that 1/a+ib= c+id

Thepotemich [5.8K]

$\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}$

Explanation

$\frac{1}{a+bi}=c+di$

Step 1

multiplicate by the conjugate

$\begin{gathered} \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(bi)^2} \\ \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(-b^2)}=\frac{a-bi}{a^2+b^2} \end{gathered}$

notice that

$\begin{gathered} \frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i \\ \frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i=c+di \\ so \\ \end{gathered}$ $\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}$

I hope this helsp you

6 0

11 months ago

Im not sure of how to solve these,any help? : 2x-5>x-2

777dan777 [17]

$2x-5>x-2\ \ \ \ \ \ \ \ \ |unknowns\ on\ the\ left\\\\2x-x>-2+5\\\\x>3\\\\x\in(\infty,3)$