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3 years ago

Which of the following equations is quadratic?

Mathematics

2 answers:

Sindrei [870]3 years ago

7 0

Answer:

I think it's 2 because the quadratic equation looks like this ax^2+bx+c=0 and option 2 looks like it.

Step-by-step explanation:

For option 1: x(x^2+1)=0

x^3+1=0 ( This also does not look like the quadratic function, so it couldn't be the answer)

For option 2: (x+3)(x+4)=5

x^2+3x+3x+12=5

x^2+6x+12=5 ( This does look like the quadratic function, so it could be the answer)

For option 3:

5(4x+2)=3

20x+10=3 ( This also does not look like the quadratic function, so it couldn't be the answer)

snow_lady [41]3 years ago

3 0

Answer:

the first one

x(x2 + 1)= 0

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The base can't be negative, therefore $x=3$ .

$3\cdot x^0+3\cdot x^1+4\cdot x^0-4\cdot x^1+0\cdot x^0=0\\3+3x+4-4x=0\\x=7$

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Step-by-step explanation:

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3 years ago

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LUCKY_DIMON [66]

516 is the dividend, 3 is the divisor, 172 would be the quotient.

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172 times 3 is 516

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3 years ago

Find the vertical asymptotes. 2x2 + 7x + 6 y = 3x2 + 10x - 8 * = [ [?], x=

Llana [10]

Answer:

$-\frac{77}{24}$

Step-by-step explanation:

1. rewrite the equation in standard form: $4\cdot \frac{3}{2}\left(y-\left(-\frac{41}{24}\right)\right)=\left(x-\left(-\frac{3}{2}\right)\right)^2$

2. find (h,k), the vertex. the vertex is $\left(h,\:k\right)=\left(-\frac{3}{2},\:-\frac{41}{24}\right)$

3. find the 'focal length' of the parabola - the focal length is the distance between the vertex and the focus. from the vertex we can see that the focal length, p, = 3/2

4. Parabola is symmetric around the y-axis and so the asymptote is a line parallel to the x-axis, a distance p from the $\left(-\frac{3}{2},\:-\frac{41}{24}\right)$ y coordinate which is at $-\frac{41}{24}\right)$ . Set up the equation:

$y=-\frac{41}{24}-p$

5. substitute and solve:

$y=-\frac{41}{24}-\frac{3}{2}$

$y = -\frac{77}{24}$

hope this helps, ask me questions if you still don't understand.

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3 years ago