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3 years ago

14

Suppose that the function f has domain all real numbers. Determine whether each function can be classified as even or odd. Expla

in.
a) g(x)= [f(x) + f(-x)]/2
b) h(x)= [f(x) – f(-x)]/2

1 answer:

dybincka [34]3 years ago

6 0

Answer:

a.Even

b.Odd

Step-by-step explanation:

We are given that function f has domain all real numbers .

We have to find that given function is even or odd

a. $g(x)=\frac{f(x)+f(-x)}{2}$

We know that if function is even then f(x)=f(-x) and if function is odd then $f(x)\neq f(-x)$

Replace x by -x then we get

$g(-x)=\frac{f(-x)+f(x)}{2}=g(x)$

Therefore, function is even .

b. $h(x)=\frac{f(x)-f(-x)}{2}$

Replace x by -x

Then we get $h(x)=\frac{f(-x)-f(x)}{2}=-\frac{f(x)-f(-x)}{2}$

$h(x)\neq h(-x)$

Hence, function is odd.

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3 years ago

Molly owes her brother $300.She works for $10 per hour as a checker at the local grocery store.Molly plans to repay her brother

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4 years ago

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3 years ago

1. Two object accumulated a charge of

In-s [12.5K]

Answer:

The magnitude of the electric force between these two objects

will be: 181.274 N.

i.e. F $=181.274$ N

Step-by-step explanation:

As

Two object accumulated a charge of 4.5 μC and another a charge of 2.8 μC.

so

q₁ = 4.5 μC = 4.5 × 10⁻⁶ C

q₂ = 2.8 μC = 2.8 × 10⁻⁶ C

separated distance = d = 2.5 cm

Calculating the magnitude of the force between two charged objects using the formula:

$F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}$

$=\:\frac{4.5\times \:\:10^{-6}\:\times \:\:2.8\:\times \:\:10^{-6}}{4\:\times \:\left(3.14\right)\:\times \:\left(8.85\times \:10^{-12}\right)\times \left(2.5\times \:\:10^{-2}\right)^2}$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:4\times \:8.85\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $4.5\times \:10^{-6}\times \:2.8\times \:10^{-6}=10^{-12}\times \:12.6$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:35.4\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $\mathrm{Multiply\:the\:numbers:}\:4\times \:8.85=35.4$

$\mathrm{Cancel\:the\:common\:factor:}\:10^{-12}$

$=\frac{12.6}{35.4\pi \left(10^{-2}\times \:2.5\right)^2}$

$=\frac{12.6}{0.025^2\times \:35.4\pi }$ ∵ $\left(10^{-2}\times \:2.5\right)^2=0.025^2$

$=\frac{12.6}{0.022125\pi }$ ∵ $35.4\pi\times 0.025^2=0.022125\pi$

$=\frac{12.6}{0.06950 }$

F $=181.274$ N

Therefore, the magnitude of the electric force between these two objects will be: 181.274 N.

i.e. F $=181.274$ N

7 0

3 years ago