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3 years ago

What is 4x squared minus x squared

Mathematics

2 answers:

kvasek [131]3 years ago

7 0

Answer:

3x^2

Step-by-step explanation:

OLEGan [10]3 years ago

3 0

Answer:

3x^2 or 3x squared

Step-by-step explanation:

4x^2-x^2=3x^2

4x squared-x squared equals x squared

when subtracting squared it's like subtracting normal numbers

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andriy [413]

The answer is f(x)=x²+2x when evaluated with -3 gives you the value of 3

Let's check all functions.
1. The function f(x)=x²+2x when evaluated with 3 gives you the value of 3:
Evaluated with x means that x = 3.
f(3) = 3² + 2 * 3 = 9 + 6 = 15
15 ≠ 3, so, this is not correct.

2. f(x)=x²-3x when evaulated with -3 give you the value of 3
Evaluated with -3 means that x = -3.
(f-3) = (-3)² - 3 * (-3) = 9 + 9 = 18
18 ≠ 3, so, this is not correct.

3. f(x)=x²+2x when evaluated with -3 gives you the value of 3
 Evaluated with -3 means that x = -3.
f(-3) = (-3)² + 2 * (-3) = 9 - 6 = 3
3 = 3, so this is correct.

4. f(x)=x²-3x when evaluated with -3 gives you the value of 3
Evaluated with 3 means that x = 3.
f(3) = (3)² - 3 * 3 = 9 - 9 = 0
0 ≠ 3, so this is not correct.

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The eccentricity e of an ellipse is defined as the number c/a, where a is the distance of a vertex from the center and c is the

Anna71 [15]

Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as $e=\frac{c}{a}$

a) Eccentricity close to 0

An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)

$\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)$

b) Eccentricity =5

$5=\frac{c}{a} \:c=5a$

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

$If\:e=\frac{c}{a} \:then\:c>0 , and\: c>0 \:then \:1>e>0$

c) Eccentricity close to 1

In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.

$a=c\\\\a^2=b^2+c^2\:(In \:the\:Pythagorean\:Theorem\: we \:should\:conceive \:b=0)$

$Then:\\\\a=c\\e=\frac{c}{a}\therefore e=1$

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