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3 years ago

The ratio of X's to O's is 15 over 10. . Use the image below to determine another ratio for X's to O's:

Mathematics

2 answers:

DENIUS [597]3 years ago

5 0

3 over 2 is another ratio for X's to O's.

iris [78.8K]3 years ago

5 0

Answer:

the answer is 3/2

Step-by-step explanation:

i got it right on my test!

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Write a quadratic function with two real zeros.

satela [25.4K]

The formula for quadratic equation is given by:

x² - (sum)x + (product) = 0

so if we have real roots of 3 and 5

<span> x² - (3 + 5)x + (3*5) = 0
</span> x² - 8<span>x + 15 = 0
</span>
so if we have real roots of -3 and -6

<span> x² - (-3 - 6)x + (-2*-6) = 0
</span> x² - (-9<span>x) + 12 = 0
</span> x² + 9<span>x + 12 = 0</span>

5 0

3 years ago

Point A is located at negative 5 over 8 and point B is located at negative 2 over 8. What is the distance between points A and B

ki77a [65]

<h2>Answer:</h2>

Option: A is the correct answer.

A. negative 5 over 8 minus negative 2 over 8 = negative 3 over 8; therefore, the distance from A to B is absolute value of negative 3 over 8 equals 3 over 8 units.

<h2>Step-by-step explanation:</h2>

The location of Point A is : A(-5/8)

and Point B is located at : B(-2/8)

We know that when two points lie over a number line then the distance between the two points is the absolute difference between the two points.

If A is located at a.

and B is located at b.

Then the distance between A and B is given by:

|a-b|

Hence, here the distance between A and B is given by:

$|\dfrac{-5}{8}-(\dfrac{-2}{8})|=|\dfrac{-5}{8}+\dfrac{2}{8}|=|\dfrac{-3}{8}|=\dfrac{3}{8}\\\\i.e.\\\\\text{Distance between A and B is}=\dfrac{3}{8}\ \text{units}$

8 0

3 years ago

Read 2 more answers

What is 2/8 x 4/5<br> a.6/40<br> b.8/40<br> c.6/13<br> d.8/13

Leya [2.2K]

The answer is B 8/40

5 0

3 years ago

Read 2 more answers

Twice the difference of x and 5 is -33

Elden [556K]

2*(x-5) = -33, so x-5 = -16.5, so x = -11.5

This is assuming that "the difference between a and b" is a-b, which seems to be the accepted interpretation.

3 0

3 years ago

The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

3 0

3 years ago