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3 years ago

How do I delete my questions?

Mathematics

1 answer:

pishuonlain [190]3 years ago

3 0

Email them ..they will help you

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40points!!!! Describe how the graph of the function is related to the graph of f(x) + x^2

muminat

Answer:

g(x) is f(x) translated downward by 7 units

Step-by-step explanation:

Adding a value to a function translates its graph upward by that value. Here, -7 is added, so the graph is translated downward by 7 units.

_____

This should be no mystery. When (x, y) is plotted on a graph, the y-coordinate specifies the vertical distance from the x-axis. Then (x, y-7) would specify a vertical distance that is 7 units less. That is, the point would be 7 units lower than the point (x, y) on the graph.

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3 years ago

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What is the equation of the line that passes through the point (-2, 6) and has a slope of -3?

poizon [28]

Answer:

y=-3x

Step-by-step explanation:

3 0

3 years ago

...........................................,<br>

irakobra [83]

Answer:

.____. whale there be no question here?

Step-by-step explanation:

.___. whale is always the answer

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3 years ago

What do you think it means when the change is positive? Negative? in a mathematical way

SashulF [63]

Answer:

A number is positive if it is greater than or equal to zero. A number is negative if it is less than or equal to zero.

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3 years ago

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Let <img src="https://tex.z-dn.net/?f=%5Calpha" id="TexFormula1" title="\alpha" alt="\alpha" align="absmiddle" class="latex-form

Margarita [4]

Answer:

$|\alpha| = 2$

Step-by-step explanation:

Since $\alpha$ and $\beta$ are complex conjugates, let's define them as follows:

$\alpha = a+bi$

$\beta = a-bi$

$\frac{\alpha}{\beta^2}=\frac{a+bi}{a^2-b^2-2abi} =\frac{(a+bi)*(a^2-b^2+2abi)}{(a^2-b^2-2abi)*(a^2-b^2+2abi)} =\frac{a^3-3ab^2+(3a^2b-b^3)i}{a^4+2a^2b^2+b^4}$

Since $\frac{\alpha}{\beta^2}$ is a real number, complex part of above result must be zero.

$3a^2b-b^3=0$

From to hold above equality, $b=0$ or $b^2=3a^2$ .

However, since $|\alpha-\beta|=2\sqrt 3$ , $b\neq 0$

So, $b =\sqrt 3 a$ or $b =-\sqrt 3 a$

And since $\alpha$ and $\beta$ are complex conjugates, taking plus or minus sign as found above will not affect the result, so let's write the last version of $\alpha$ and $\beta$ as follows:

$\alpha = a+\sqrt 3 ai$

$\beta = a-\sqrt 3 ai$

Since $|\alpha-\beta|=2\sqrt 3$

$|a+\sqrt 3 ai-a+\sqrt 3 ai|=2\sqrt 3$ ⇒ $a = 1$

Finally, $\alpha = 1+\sqrt 3 i$ ⇒ $|\alpha| = \sqrt{(1^2+\sqrt 3^2)}=2$

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3 years ago