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2 years ago

HELP ME PLEASE !!!! ill be marking brainliest

Mathematics

1 answer:

RUDIKE [14]2 years ago

5 0

Using the concepts of even and odd functions, the points that will be contained in the function are given as follows:

Even: (-2,-5).

Odd: (-2,5).

<h3>What are even and odd functions?</h3>

Even and odd functions both have relationships between f(x) and f(-x), as follows:

For even functions, f(-x) = f(x).

For odd functions, f(-x) = -f(x).

This function contains point (2,-5), that is, f(2) = -5, hence:

For an even function, f(-2) = -5, hence it contains point (-2, -5).

For an odd function, f(-2) = 5, hence it contains point (-2, 5).

More can be learned about even and odd functions at brainly.com/question/24410059

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What’s the linear equation of this graph? PLEASE HELP

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Answer:

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Step-by-step explanation:

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3 years ago

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Which expression equivalent to k^1/4 · cubed root of k^2

mina [271]

From the info i'm getting, the best answer is ought to be B, because of increase in the denominator & it's numerator! It keeps the same pace as 1/4.

I do recommend to double check this with a classmate!

If not B, then my best guess is A.

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3 years ago

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sammy [17]

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3 years ago

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Show that a = −1 + √3i and b = 2 satisfy 1/a+b=1/a + 1/b

Zarrin [17]

Answer:

LHS = $\frac{1 - \sqrt3i}{4}$ = RHS = $\frac{1 - \sqrt3i}{4}$

Step-by-step explanation:

Data provided in the question:

a = −1 + √3i and b = 2

to prove:

$\frac{1}{a+b}=\frac{1}{a} + \frac{1}{b}$

Considering the LHS

⇒ $\frac{1}{a+b}$

substituting the value of a and b, we get

⇒ $\frac{1}{−1 + \sqrt3i+2}$

⇒ $\frac{1}{1 + \sqrt3i}$

on multiplying and dividing by conjugate ( 1 - √3i )

we get

$\frac{1}{1 + \sqrt3i}\times\frac{1 - \sqrt3i}{1 - \sqrt3i}$

$\frac{1 - \sqrt3i}{(1^2 - (\sqrt3i)^2}$

$\frac{1 - \sqrt3i}{1 + 3}$ (as (√i)² = -1 )

$\frac{1 - \sqrt3i}{4}$

Now,

considering the RHS

$\frac{1}{a} + \frac{1}{b}$

substituting the value of a and b, we get

⇒ $\frac{1}{-1 + \sqrt3i} + \frac{1}{2}$

⇒ $\frac{2\times1 + ( -1 + \sqrt3i)\times1}{(-1 + \sqrt3i)\times2}$

⇒ $\frac{2 + ( -1 + \sqrt3i)}{(-1 + \sqrt3i)\times2}$

⇒ $\frac{1 + \sqrt3i}{(-1 + \sqrt3i)\times2}$

now,

on multiplying and dividing by conjugate ( -1 - √3i )

we get

$\frac{1 + \sqrt3i}{(−1 + \sqrt3i)\times2}\times\frac{-1 - \sqrt3i}{-1 - \sqrt3i}$

$\frac{1 + \sqrt3i}{(−1 + \sqrt3i)\times2}\times\frac{-1( 1 + \sqrt3i)}{-1 - \sqrt3i}$

$\frac{(1 + \sqrt3i}^2\times(-1){((-1)^2 - (\sqrt3i)^2)\times2}$

$\frac{(1^2 + (\sqrt3i)^2+2(1)(\sqrt3i)\times(-1)}{(1 + 3)\times2}$

$\frac{(1 - 3 + 2\sqrt3i)\times(-1)}{(4)\times2}$

$\frac{(-2 + 2\sqrt3i)\times(-1)}{(4)\times2}$

$\frac{-2( 1 - 2\sqrt3i)\times(-1)}{(4)\times2}$

$\frac{( 1 - 2\sqrt3i)}{(4)}$

Since, LHS = RHS

hence satisfied

3 0

3 years ago

There are 96 new houses being built in a neighborhood. Last month,1 over 4 of them were sold. This month, 1 over 3 of the remain

Elis [28]

1/4 of 96 were sold... replace "of" with ×

1/4 × 96 = 96/4 = 24

So the first month 24 houses were sold so that leaves...

96 - 24 = 72 houses.

1/3 of 72 were sold the next month

1/3 × 72 = 72/3 = 24

24 houses were sold the second month so there are...

72 - 24 = 48 houses left

8 0

3 years ago