All categories

1 year ago

Please help, need for math hw and cant figure it out

Mathematics

1 answer:

swat321 year ago

7 0

The standard form of the quadratic function f(x) = -3x^2 + 6x - 2 is f(x) = -3x^2 + 6x - 2

<h3>How to represent the quadratic function in standard form?</h3>

The quadratic function is given as

f(x) = -3x^2 + 6x - 2

The standard form of a quadratic function is represented as:

f(x) = ax^2 + bx + c

When both equations are compared, we can see that the function f(x) = -3x^2 + 6x - 2 is already in standard form

Where

a = -3

b = 6

c = -2

Hence, the standard form of the quadratic function f(x) = -3x^2 + 6x - 2 is f(x) = -3x^2 + 6x - 2

Read more about quadratic function at

brainly.com/question/25841119

#SPJ1

You might be interested in

PLS HELP I NEED ANSWER ASAP!<br> PPLS BE RIGHT!

Luba_88 [7]

Answer:

They used 6 vans and 3 buses

Step-by-step explanation:

x + y= 9

13x + 40y= 198

x= 9 - y

13(9 - y) + 40y= 198

117 - 13y + 40y= 198

27y= 81

y= 3

9-3= 6

4 0

3 years ago

Read 2 more answers

What is the value of n in this equation?<br>n+ 134 = 379<br>423<br>245<br>334

alexira [117]

Answer:

the answer to this question is 245

Step-by-step explanation:

n+134=379

-134 -134

n=245

5 0

3 years ago

Read 2 more answers

How many solutions does 2.3y+3.2-y=2.1+1.3y+1.1 have?

jeka57 [31]

1.3y +3.2 = 1.3y +3.2

we notice that both sides of the equation are identical. So substitute y with any value and the equation would still be correct.

y has infinite solutions.

6 0

3 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

3 years ago

Solve. -5 3/4-3 1/2=?

g100num [7]

Answer: -9 1/4.

Step-by-step explanation:

6 0

2 years ago