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3 years ago

11

Consider what you know about the relative locations of the vertex and x-intercepts of the graph of a quadratic function. Describ

e in detail how the quadratic formula defines these points algebraically.

1 answer:

Bas_tet [7]3 years ago

4 0

Hello,

<span>the relative locations of the vertex is (a,b)
and x-intercepts is x₀

y=k(x-a)+b
and
0=k(x₀-a)+b==>k=-b/(x₀-a)

==>y=-b/(x₀-a)*(x-a)+b
</span>

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Solve (2/3)-(1/x)=5/6

Keith_Richards [23]

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$12 - \frac{18}{x} = 15$ Multiply both sides by x
12x - 18 = 15x Subtract 12x from both sides
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-6 = x Switch the sides to make it easier to read
x = -6

Check your answer by plugging -6 in for the x in the original problem:

$\frac{2}{3} - \frac{1}{x} = \frac{5}{6}$ Plug in -6 for x
$\frac{2}{3} - \frac{1}{-6} = \frac{5}{6}$ Change $\frac{2}{3}$ to $\frac{4}{6}$ so all denominators are the same
$\frac{4}{6} - \frac{1}{-6} = \frac{5}{6}$ Change $\frac{1}{-6}$ to $\frac{-1}{6}$ for easier work
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3 years ago

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Answer:

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Step-by-step explanation:

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lbvjy [14]

Given:

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$(2x+3y)(2x-3y)=4x^2+6y^2$

To find:

The error in the given equation and correct it.

Solution:

We have,

$(2x+3y)(2x-3y)=4x^2+6y^2$

Taking left-hand side, we get

$L.H.S.=(2x+3y)(2x-3y)$

$L.H.S.=(2x)^2-(3y)^2$ $[\because a^2-b^2=(a-b)(a+b)]$

$L.H.S.=(2)^2(x)^2-(3)^2(y)^2$ $[\because (ab)^x=a^xb^x]$

$L.H.S.=4x^2-9y^2$

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Free_Kalibri [48]

Answer:

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