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2 years ago

Please help with POWER FUNCTIONS!!!!!

Mathematics

1 answer:

grigory [225]2 years ago

7 0

First answer, both functions are parabolas (bowl shaped) that curve upward. The 3 makes (1/3)f(x) = g(x) the graph narrower.

36 = a(3)^2
36 = 9a
36/9 = a
4 = a

f(x) = 4x^2

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2 years ago

Select the correct answer.

vagabundo [1.1K]

Answer

D. `(y+4) = (5)/(3)(x+ 1)`

Step by step explanation

The given points are (-1, -4) and (2, 1)

Slope = (y2 - y1)/(x2 - x1)

= (1 - (-4)) / (2 - (-1))

= (1 + 4) / (2 +1)

m = 5/3

When two points are given, the equation of the line is

(y - y1) = m (x - x1)

y -(-4) = 5/3 (x - (-1))

y + 4 = 5/3 (x + 1)

The answer is (y+4) = (5)/(3)(x+ 1)

Thank you.

5 0

2 years ago

Read 2 more answers

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

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1 year ago