Ask question

Ask question

All categories

3 years ago

5

Identifying the values a, b, and c is the first step in using the Quadratic Formula to find solution(s) to a quadratic equation.

What are the values a, b, and c in the following quadratic equation? −6x2 = −9x + 7

2 answers:

lesya692 [45]3 years ago

3 0

The values of a, b, and c in given quadratic equation are:

a = 6 and b = -9 and c = 7

<em><u>Solution:</u></em>

Given quadratic equation is:

$-6x^2 = -9x + 7$

Let us first convert the given quadratic equation to standard form

The standard form is $ax^2+bx+c=0$ with a, b, and c being constants, or numerical coefficients, and x is an unknown variable

$-6x^2 = -9x + 7\\\\-6x^2+9x-7=0\\\\6x^2-9x+7=0$

Now we have to find the values of a, b, c

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Thus comparing $6x^2-9x+7=0$ with standard form of quadratic equation $a x^{2}+b x+c=0$

a = 6

b = -9

c = 7

Thus values of a, b, and c are found

grin007 [14]3 years ago

3 0

Answer:

a=-6 b=-9 c=7

i just took this test and i got it right

You might be interested in

Juliet needs to rewrite 3(x+2)=6y in slope intercept form so she can easily graph it which step would be correct to start the pr

otez555 [7]

<span>B:she could divided both sides by 3 to get x+2=2y</span>

6 0

4 years ago

Read 2 more answers

Where y=5x-7 and -3x-2y=-12 Write your answer as x,y

Mariulka [41]

Answer:

(x,y) = (2,3)

Step-by-step explanation:

y=5x-7 (i)

-3x-2y= -12

-(3x+2y)= -12

3x+2y= 12 (ii)

By putting the value of y from (i) in (ii)

3x+2(5x-7)=12

3x+10x-14=12

13x=12+14

13x= 26

x = $\frac{26}{13}$

x= 2

By putting the value of x in (i)

y= 5(2)-7

y= 10-7

y= 3

(x,y) = (2,3)

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%2By%3D1%7D%20%5Catop%20%7Bx-2y%3D4%7D%7D%20%5Cright.%20%5C%5C%5Clef

brilliants [131]

Answer:

<em>(a) x=2, y=-1</em>

<em>(b) x=2, y=2</em>

<em>(c)</em> $\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

<em>(d) x=-2, y=-7</em>

Step-by-step explanation:

<u>Cramer's Rule</u>

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

$\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.$

We call the determinant of the system

$\Delta=\begin{vmatrix}a &b \\c &d \end{vmatrix}$

We also define:

$\Delta_x=\begin{vmatrix}p &b \\q &d \end{vmatrix}$

And

$\Delta_y=\begin{vmatrix}a &p \\c &q \end{vmatrix}$

The solution for x and y is

$\displaystyle x=\frac{\Delta_x}{\Delta}$

$\displaystyle y=\frac{\Delta_y}{\Delta}$

(a) The system to solve is

$\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.$

Calculating:

$\Delta=\begin{vmatrix}1 &1 \\1 &-2 \end{vmatrix}=-2-1=-3$

$\Delta_x=\begin{vmatrix}1 &1 \\4 &-2 \end{vmatrix}=-2-4=-6$

$\Delta_y=\begin{vmatrix}1 &1 \\1 &4 \end{vmatrix}=4-3=3$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1$

The solution is x=2, y=-1

(b) The system to solve is

$\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.$

Calculating:

$\Delta=\begin{vmatrix}4 &-1 \\1 &-1 \end{vmatrix}=-4+1=-3$

$\Delta_x=\begin{vmatrix}6 &-1 \\0 &-1 \end{vmatrix}=-6-0=-6$

$\Delta_y=\begin{vmatrix}4 &6 \\1 &0 \end{vmatrix}=0-6=-6$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2$

The solution is x=2, y=2

(c) The system to solve is

$\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.$

Calculating:

$\Delta=\begin{vmatrix}-1 &2 \\1 &2 \end{vmatrix}=-2-2=-4$

$\Delta_x=\begin{vmatrix}0 &2 \\5 &2 \end{vmatrix}=0-10=-10$

$\Delta_y=\begin{vmatrix}-1 &0 \\1 &5 \end{vmatrix}=-5-0=-5$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}$

The solution is

$\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) The system to solve is

$\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.$

Calculating:

$\Delta=\begin{vmatrix}6 &-1 \\4 &-2 \end{vmatrix}=-12+4=-8$

$\Delta_x=\begin{vmatrix}-5 &-1 \\6 &-2 \end{vmatrix}=10+6=16$

$\Delta_y=\begin{vmatrix}6 &-5 \\4 &6 \end{vmatrix}=36+20=56$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7$

The solution is x=-2, y=-7

4 0

3 years ago

1. Which of the following points is a solution of y > |x| + 5?

Dennis_Churaev [7]

1. iThis is shaped like a V with its vertex at the point (0,5)
( its ( 1,7)

2. -3 <= -1-1 = =2

so its (-1, -3)

7 0

3 years ago

Read 2 more answers

It was 4:00 P.M. when John started mowing the lawn and it took him 50 minutes to finish. What time did John finish mowing the la

Oksi-84 [34.3K]

Answer:

4:50 P.M.

Step-by-step explanation:

Sorry I don't know how to properly explain it but I just added 50 minutes to 4 o' clock to get the answer.

Hope this helps!

Maybe brainliest?

7 0

3 years ago

Read 2 more answers