All categories

3 years ago

Create a quadratic equation that can be solved by factoring. Give the factored answer.

1 answer:

3 0

Quadratic eq standard form is

Ax^2+bx+c=0

<span>a, b and c are known values.
a can't be 0.</span><span>"x" is the variable

lets take an example

</span><span>x^2 − 4x+2 = 0
</span><span>
x^2-2x-2x+2=0
x(x-2) - 1(x-2)=0

x=1,2

</span>

You might be interested in

-3(8n+10) <br> distributive property

Lerok [7]

-24n - 30, i suggest using math.way before asking a math question aha

6 0

2 years ago

Read 2 more answers

A rectangle has an area of 75.2m

Degger [83]

Answer:

94.8

Step-by-step explanation:

Aera is : length *width

75.2 = 0.8 * x then x =94

Perimeter is : length + width =94 +0.8

3 0

2 years ago

quadrilateral paid is a rectangle whose diagonals have the endpoints p(-3, -2)i(4, -7) and a(4, -2)d(-3, -7). find the diagonals

julia-pushkina [17]

Answer:

The intersection point of diagonals is (0.5,-4.5).

Step-by-step explanation:

The end points of first diagonal are p(-3, -2) and i(4, -7).

The end points of second diagonal are a(4, -2) and d(-3, -7).

If a line passing through two points, then the equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

The equation of first diagonal is

$y-(-2)=\frac{-7-(-2)}{4-(-3)}(x-(-3))$

$y+2=\frac{-5}{7}(x+3)$

$7y+14=-5(x+3)$

$7y+14=-5x-15$

$5x+7y=-29$ .... (1)

The equation of second diagonal is

$y-(-2)=\frac{-7-(-2)}{-3-4}(x-4)$

$y+2=\frac{-5}{-7}(x-4)$

$7y+14=5(x-4)$

$7y+14=5x-20$

$5x-7y=34$ .... (2)

Add equation (1) and (2),

$10x=5$

$x=0.5$

Put this value in (1).

$5(0.5)+7y=-29$

$y=-4.5$

The intersection point of diagonals is (0.5,-4.5).

3 0

3 years ago

A basketball team has 10 players. Before each game, the coach picks 2 of those players to carry the team's

labwork [276]

The different group of 2 players that the coach can pick is 45 groups.

<h3>Selection of groups</h3>

The selection of groups of 2 player can be done using the method of combination.

<h3>Different groups of 2 players</h3>

The different group of 2 players that the coach can pick from the 10 players is calculated as follows;

n = 10C2

$n = \frac{10!}{(10-2)! 2!} = \frac{10 \times 9\times 8!}{8! \times 2} = 45$

Thus, the different group of 2 players that the coach can pick is 45 groups.

Learn more about combinations here: brainly.com/question/25821700

6 0

2 years ago

Multiply (x-2)(3x+4) using the foil method

masya89 [10]

<span>Multiply (x-2)(3x+4) using the foil method

</span> $(x-2)(3x+4) = \\ \\ (x*3x) +(x*4) +(-2*3x) +(-2*4) = \\ \\ 3x^2 + 4x - 6x - 8 = \\ \\ 3x^2 -2x - 8 \\ \\ \\ \boxed{(x-2)(3x+4) =3x^2 -2x - 8}$ <span>

</span>

4 0

3 years ago