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3 years ago

8

What is the difference between a linear equation and a quadratic equation?(in 7 sentences)

1 answer:

Kaylis [27]3 years ago

7 0

A linear equation in two variables doesn't involve any power higher than one for either variable. It has the general form Ax + By + C = 0, where A, B and C are constants. It's possible to simplify this to y = mx + b, where m = ( −A / B) and b is the value of y when x = 0. A quadratic equation, on the other hand, involves one of the variables raised to the second power. It has the general form y = ax2 + bx + c. Apart from the adding complexity of solving a quadratic equation compared to a linear one, the two equations produce different types of graphs.

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A=1/4 and b=6.<br><br> −8ab=

podryga [215]

Answer:

When $a=\frac{1}{4}$ and $b=6$ :

$-8ab=-12$

Step-by-step explanation:

-8ab can be seen as -8×a×b. Insert the given values:

$-8*\frac{1}{4} *6$

Simplify multiplication from left to right:

$-8*\frac{1}{4} \\\\\frac{-8}{1} *\frac{1}{4} =\frac{-8}{4} =-2$

Insert and solve:

$-2*6=-12$

:Done

3 0

3 years ago

I need help on these please

sleet_krkn [62]

Answer:

its the third answer, fifth, and first. Im not sure tho.

Step-by-step explanation:

if it isnt right im so sorry T^T

5 0

3 years ago

Y=r + 5x + 4<br> please help!!!!

frez [133]

Your answer would be R=-5x-4+y

6 0

2 years ago

Annie is framing a photo with a length of 6 inches and a width of 4 inches. The distance from the edge of the photo to the edge

Ede4ka [16]

Answer:

Part a) The quadratic function is $4x^{2} +20x-39=0$

Part b) The value of x is $1.5\ in$

Part c) The photo and frame together are $7\ in$ wide

Step-by-step explanation:

Part a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame

Let

x----> the distance from the edge of the photo to the edge of the frame

we know that

$(6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0$

Part b) What is the value of x?

Solve the quadratic equation $4x^{2} +20x-39=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem

we have

$4x^{2} +20x-39=0$

so

$a=4\\b=20\\c=-39$

substitute in the formula

$x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}$

$x=\frac{-20(+/-)\sqrt{1,024}} {8}$

$x=\frac{-20(+/-)32} {8}$

$x=\frac{-20(+)32} {8}=1.5\ in$ -----> the solution

$x=\frac{-20(-)32} {8}=-6.5\ in$

Part c) How wide are the photo and frame together?

$(4+2x)=4+2(1.5)=7\ in$

5 0

3 years ago

Need a quick answer please for math asap!

morpeh [17]

Because the discriminant is less than zero, there are no real solutions in the equation.

4 0

3 years ago

Read 2 more answers