<img src="https://tex.z-dn.net/?f=12x%5E%7B4%7D%20%2B30x%5E%7B3%7D%20%2B18x%5E%7B2%7D" id="TexFormula1" title="12x^{4} +30x^{3}

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Rama09 [41]

3 years ago

+18x^{2}" alt="12x^{4} +30x^{3} +18x^{2}" align="absmiddle" class="latex-formula">

Factorize

Mathematics

1 answer:

nika2105 [10]3 years ago

6 0

6x squared (x + 1)(2x + 3)

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The area of a trampoline is 49 pi sqared. What's the diameter of the trampoline?

Flauer [41]

Answer:

The diameter is 14

Step-by-step explanation:

The area of a circle is given by

A = pi r^2

We know the area is 49 pi

49 pi = pi r^2

Divide each side by pi

49 pi/ pi = pi r^2/pi

49 = r^2

Take the square root of each side

sqrt(49) = sqrt(r^2)

7 =r

We want the diameter

d = 2r

d = 2*7

d = 14

7 0

3 years ago

This graph represents a quadratic function. What is the function’s equation written in factored form and in vertex form?

AnnZ [28]

Answer:

$\textsf{Factored form}: \quad f(x)=2x(x-4)$

$\textsf{Vertex form}: \quad f(x)=2(x-2)^2-8$

Step-by-step explanation:

Factored form of a quadratic function

$f(x)=a(x-p)(x-q)$

where:

p and q are the x-intercepts
a is some constant

Given x-intercepts:

(0, 0)
(4, 0)

Therefore:

$\implies f(x)=a(x-0)(x-4)$

$\implies f(x)=ax(x-4)$

To find a, substitute the given vertex (2, -8) into the equation and solve for a:

$\implies 2a(2-4)=-8$

$\implies -4a=-8$

$\implies a=2$

Therefore, the function's equation in factored form is:

$f(x)=2x(x-4)$

Vertex form of a quadratic equation

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

where:

(h, k) is the vertex
a is some constant

Given:

vertex = (2, -8)

Therefore:

$\implies f(x)=a(x-2)^2-8$

To find the constant a, substitute one of the x-intercepts into the equation and solve for a:

$\implies a(0-2)^2-8=0$

$\implies 4a-8=0$

$\implies a=2$

Therefore, the function's equation in vertex form is:

$f(x)=2(x-2)^2-8$

4 0

2 years ago

Read 2 more answers

Give me a correct answer, My answer probably wrong.

Georgia [21]

Hello!

This is a problem about the general solution of a differential equation.

What we can first do here is separate the variables so that we have the same variable for each side (ex. $dy$ with the $y$ term and $dx$ with the $x$ term).

$\frac{dy}{dx}=\frac{x-1}{3y^2}$

$3y^2dy=x-1dx$

Then, we can integrate using the power rule to get rid of the differentiating terms, remember to add the constant of integration, C, to at least one side of the resulting equation.

$y^3=\frac{1}{2}x^2-x+C$