All categories

2 years ago

A quadratic equation is shown below:

1 answer:

6 0

Part A:
3x^2-15x+20=0
ax^2+bx+c=0; a=3, b=-15, c=20
Radicand: R=b^2-4ac
R=(-15)^2-4(3)(20)
R=225-240
R=-15<0, then the equation has two imaginary solutions (no real)

Part B:
3x^2+5x-8=0
a=3, b=5, c=-8
R=(5)^2-4(3)(-8)
R=25+96
R=121>0, then the equation has two different real solutions:
x=[-b+-sqrt(R)] / (2a)
x=[-5+-sqrt(121)] / [2(3)]
x=(-5+-11) / 6

x1=(-5-11)/6=(-16)/6→x1=-8/3

x2=(-5+11)/6=6/6→x2=1

Solutions: x=-8/3 and x=1

I shose this method because I can get the solutions directly, and I don't have to guess the possible solutions

You might be interested in

Question 10 of 10

dangina [55]

Answer:

thanks for the points bye bye for now

Step-by-step explanation:

thanks for the points bye bye for now love u

3 0

2 years ago

John bought new headphones originally listed for $ 80.99 . They are \(35\%)| off. Which equation can be used to find the amount

Katarina [22]

Answer:

$28.35

Step-by-step explanation:

The headphones are $80.99 and are 35% off. To find the amount he will save, change the 35% to a decimal by moving it two times to the left, which is 0.35. Multiply 80.99 and 0.35 to get 28.3465. Round to get $28.35.

7 0

1 year ago

What is the volume of this prism?

vovikov84 [41]

Answer:

The answer is 13 1/8ft3

Explanation:

Substitute into the formula:

Volume = 3 1/2 x 2 1/2 x 1 1/2

= 7/2 x 5/2 x 3/2

= 105/8

= 13 1/8 ft3

8 0

2 years ago

Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval.

Romashka [77]

By definition, the average change of rate is given by:
$AVR = \frac{f(x2)-f(x1)}{x2-x1}$
We will calculate AVR for each of the functions.
We have then:

f(x) = x^2 + 3x interval: [-2, 3]:
$f(-2) = x^2 + 3x = (-2)^2 + 3(-2) = 4 - 6 = -2

f(3) = x^2 + 3x = (3)^2 + 3(3) = 9 + 9 = 18$
$AVR = \frac{-2-18}{-2-3}$
$AVR = \frac{-20}{-5}$
$AVR = 4$

f(x) = 3x - 8 interval: [4, 5]:
$f(4) = 3(4) - 8 = 12 - 8 = 4 f(5) = 3(5) - 8 = 15 - 8 = 7$
$AVR = \frac{7-4}{5-4}$
$AVR = \frac{3}{1}$
$AVR = 3$

f(x) = x^2 - 2x interval: [-3, 4]
$f(-3) = (-3)^2 - 2(-3) = 9 + 6 = 15

f(4) = (4)^2 - 2(4) = 16 - 8 = 8$
$AVR = \frac{8-15}{4+3}$
$AVR = \frac{-7}{7}$
$AVR = -1$

f(x) = x^2 - 5 interval: [-1, 1]
$f(-1) = (-1)^2 - 5 = 1 - 5 = -4

f(1) = (1)^2 - 5 = 1 - 5 = -4$
$AVR = \frac{-4+4}{1+1}$
$AVR = \frac{0}{2}$
$AVR = 0$

Answer:
these functions from the greatest to the least value based on the average rate of change are:
f(x) = x^2 + 3x
f(x) = 3x - 8
f(x) = x^2 - 5
f(x) = x^2 - 2x

5 0

3 years ago

Is the square root of 75/3 a rational or irrational number?

erma4kov [3.2K]

The square root of 75/3 is a irrational number

6 0

3 years ago

Read 2 more answers