Ask question

Ask question

All categories

2 years ago

13

Some Quadratic functions I desperately need some help

1 answer:

Daniel [21]2 years ago

6 0

Answer:

$\large\boxed{0=3x^2+2x-2}\\\\\boxed{0=-3x+3x^2-2}\\\\\boxed{0=-1x-2+3x^2}$

Step-by-step explanation:

$\text{The quadratic equation:}\\\\ax^2+bx+c=0\\\\a-coefficient\\\\c-constant\ term\\\\\text{We have}\ coefficient\ a=3\ \text{and}\ constant\ term\ c=-2.\\\\\text{Therefore the equation is:}\\\\3x^2+bx-2=0\\3x^2-2+bx=0\\bx+3x^2-2=0\\bx-2+3x^2=0\\-2+3x^2+bx=0\\-2+bx+3x^2=0$

You might be interested in

A can of juice is 6 inches high, and its base has a diameter

xenn [34]

Answer:

24 pie, or, 75.36 inches cubed

Step-by-step explanation:

The formula for volume of a cylinder is pie*radius^2*height

The radius is half the diameter, so the radius is 2 inches.

Plug in given information into the formula and solve.

4 0

2 years ago

I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

Rudik [331]

Answer:

a=1/2b, 2a=b

Step-by-step explanation:

Since angle a is 1/2 the size of angle b, a=1/2b and b=2a. In order to get a numeral answer, you need the number of one of the angles.

7 0

2 years ago

Pleeeeeeeeaaaseeeeee help meeeeeeee!!!!!!!!!

zaharov [31]

Answer:

Option D.

$A =96\pi\ cm^2$

Step-by-step explanation:

The area of the circular bases is:

$A_c = 2\pi(a) ^ 2$

Where

$a=4\ cm$ is the radius of the circle

Then

$A = 2\pi(4) ^ 2$

$A = 32\pi\ cm^2$

The area of the rectangle is:

$A_r=b * 2\pi r$

Where

$b=8\ cm$

b is the width of the rectangle and $2\pi r$ is the length

Then the area of the rectangle is:

$A_r=8 * 2\pi (4)$

$A_r=64\pi\ cm^2$

Finally the total area is:

$A = A_c + A_r\\\\A = 32\pi\ cm^2 + 64\pi\ cm^2\\\\$

$A =96\pi\ cm^2$

7 0

3 years ago

Read 2 more answers

Simplify: 27k^5m^8/(4k^3)(9m^2)

Yakvenalex [24]

The answer to your question is 3k^2m^6/4

4 0

2 years ago

In this line of circles, the ratio of red circles to blue circles is 1:3 How many more circles would need to be added to make th

zaharov [31]

Answer:

The number of red circles to be added would be equal to 8 times the original amount of red circles

Step-by-step explanation:

Let

x ----> the number of red circles

y ----> the number of blue circles

Originally

$\frac{x}{y}=\frac{1}{3}$

$y=3x$ ----> equation A

How many more circles would need to be added to make the ratio of red to blue 3:1

Let

n ----> the number of additional red circles

$\frac{x+n}{y}=\frac{3}{1}$ ----> equation B

substitute equation A in equation B

$\frac{x+n}{3x}=\frac{3}{1}$

solve for n

$x+n=9x\\n=9x-x\\n=8x$

therefore

The number of red circles to be added would be equal to 8 times the original amount of red circles

5 0

3 years ago