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2 years ago

Which equation is y = â€""6x2 3x 2 rewritten in vertex form?.

Mathematics

1 answer:

Ber [7]2 years ago

6 0

The vertex form needed is :

$-6(x-0.25)^2 +2.375$

Given the equation:

$-6x^2 + 3x + 2\\$

The vertex form of $f(x) = ax^2 + b(x) + c$ is given by:

$f(x) = a(x-h)^2 + k$

Here, (h,k) is the vertex of the parabola equation given.

Conversion will go like this:

$-6x^2 + 3x + 2\\= -6(x^2 - 0.5x) + 2\\= -6(x^2 -0.5x + 0.0625 - 0.0625) + 2\\= -6(x^2 -0.5x + 0.0625) + 2 +0.375\\= -6(x-0.25)^2 + 2.375\\\\$

This is the resultant vertex form.

Thus the vertex form needed is :

$-6(x-0.25)^2 +2.375$

Learn more here:

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Which statement about isosceles triangles is true?

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2 years ago

One way to build stairs is to attach triangular blocks to an angled support, as shown at the right. If the support makes a 32Â°

Sauron [17]

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Step-by-step explanation:
Since we have given that
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2 years ago

Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx = 1$

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have

$\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx = 1$

Since c is a constant, we can bring it out of the integral sign; to give us

$c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx = 1$

Open the bracket

$c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx = 1$

Integrate with respect to y

$c\int\limits^3_0 {\frac{xy^{2}}{2} +\frac{xy^{3}}{3} } \, dx (0,3}) = 1$

Substitute 0 and 3 for y

$c\int\limits^3_0 {(\frac{x* 3^{2}}{2} +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2} +\frac{x * 0^{3}}{3})} \, dx = 1$

$c\int\limits^3_0 {(\frac{x* 9}{2} +\frac{x * 27}{3} ) - (0 +0) \, dx = 1$

$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

$c\int\limits^3_0 {\frac{81x}{6} \, dx = 1$

Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

$\frac{c}{6} * \frac{81 * 3^{2} - 81 * 0^{2}}{2} = 1$

$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

8 0

2 years ago

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Eddi Din [679]

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2 years ago

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malfutka [58]

Answer:

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1 year ago