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3 years ago

Can anyone answer this ?

1 answer:

8 0

$(3 {x}^{2} - 4x - 1) + (8 {x}^{2} - x + 6) \\ \\ combining \: the \: like \: terms,\:we\:have\\ \\ = 3 {x}^{2} + 8 {x}^{2} - 4x - x - 1 + 6 \\ \\ = 11 {x}^{2} - 5x + 5$

$(6 {x}^{2} - x - 4) + ( 2{x}^{2} + 5x - 5) \\ \\combining \: the \: like \: terms, \: we \: have \\ \\= 6 {x}^{2} + 2 {x}^{2} - x + 5x - 4 - 5 \\ \\= 8 {x}^{2} + 4x - 9$

$(4 {x}^{2} + 2x - 5) + (6 {x}^{2} - x - 5) \\ \\ combining \: th e\:l ike \: terms, \: we \: have \\ \\= 4 {x}^{2} + 6 {x}^{2} + 2x - x - 5 - 5 \\ \\= 10 {x}^{2} +x - 10$

$\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{.}}}}}$

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2 years ago

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

andrezito [222]

Answer:

Given the equation: $3x^2+10x+c =0$

A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

Simplify:

$x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}$

Also, it is given that the difference of two roots of the given equation is $4\frac{2}{3} = \frac{14}{3}$

i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

$\sqrt{100- 12c} = 14$

Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

3 0

2 years ago

Consider the equality xy k. Write the following inverse proportion: y is inversely proportional to x. When y = 12, x=5.

skelet666 [1.2K]

Answer:

$y=\dfrac {60} {x}$ or $xy=60$ (depending on your teacher's format preference)

Step-by-step explanation:

<h3><u>Proportionality background</u></h3>

Proportionality is sometimes called "variation". (ex. " 'y' varies inversely as 'x' ")

There are two main types of proportionality/variation:

Direct
Inverse.

Every proportionality, regardless of whether it is direct or inverse, will have a constant of proportionality (I'm going to call it "k").

Below are several different examples of both types of proportionality, and how they might be stated in words:

$y=kx$ y is directly proportional to x
$y=kx^2$ y is directly proportional to x squared
$y=kx^3$ y is directly proportional to x cubed
$y=k\sqrt{x}}$ y is directly proportional to the square root of x
$y=\dfrac {k} {x}$ y is inversely proportional to x
$y=\dfrac {k} {x^2}$ y is inversely proportional to x squared

From these examples, we see that two things:

things that are <u>directly proportional</u> -- the thing is <u>multipli</u>ed to the constant of proportionality "k"
things that are <u>inversely proportional</u> -- the thing is <u>divide</u>d from the constant of proportionality "k".

<h3><u>Looking at our question</u></h3>

In our question, y is inversely proportional to x, so the equation we're looking at is the following $y=\dfrac {k} {x}$ .

It isn't yet clear what the constant of proportionality "k" is for this situation, but we are given enough information to solve for it: "When y=12, x=5."

We can substitute this known relationship pair, and find the "k" that relates this pair of numbers:

<h3><u>Solving for k, and finding the general equation</u></h3>

General Inverse variation equation...

$y=\dfrac {k} {x}$

Substituting known values...

$(12)=\dfrac {k} {(5)}$

Multiplying both sides by 5...

$(12)*5= \left ( \dfrac {k} {5} \right ) *5$

Simplifying/arithmetic...

$60=k$

So, for our situation, k=60. So the inverse proportionality relationship equation for this situation is $y=\dfrac {60} {x}$ .

The way your question is phrased, they may prefer the form: $xy=60$

7 0

1 year ago