contestada

ANYONE PLEASE HELP
# BRAINLIEST FOR HELP
1) In the quadratic equation, k represents a real number.
3x^2+2kx+2=0
Which value of k gives imaginary solutions to the equation? Select all the values that apply.
A. − 5
B. − 3
C. − 1
D. 2
E. 3
F. 4
2) 1/2 x+12=8+4( x+1)
Which could be a correct first step to solve the equation? Select all the steps that apply.
A. Distribute 4 to the quantity ( x+1).
B. Add − 8 to each side of the equation.
C. Distribute 12 to the quantity ( x+1).
D. Multiply each side of the equation by 2.
E. Subtract 12 from each side of the equation.
F. Add together the 8 and the 4 on the right side of the equation.
G. Multiply the left side of the equation by 2 to eliminate the fraction.

3) In the equation of f(x) shown, r and s are positive real numbers.
f(x)=x^2−2(r−s)x+(r−s)^2+(r+s)
Select the options from the drop-down menus to correctly rewrite f(x).
FILL IN: f(x)= ( ......... ......... )2 ..........

4) A city’s population, P, changes according to the function P( x )=50,000 ( 0.95 )^x. where x represents the number of years after the year 2000.

Select the options from the drop-down menus to complete an accurate description of the population.

The population of the city in the year 2000 was(...……..) . The population is (..............) by (..........)each year.

Respuesta :

raphealnwobi

Answer:

1)

C. − 1

D. 2

2)

A. Distribute 4 to the quantity ( x+1).

B. Add − 8 to each side of the equation.

D. Multiply each side of the equation by 2

E. Subtract 12 from each side of the equation.

3)

[tex]f(x)=(r-x-s)^2+(r+s)[/tex]

4) The population of the city in the year 2000 was 50000. The population is decreases by 0.95 each year

Step-by-step explanation:

a) Given an equation:

ax² + bx + c = 0

The equation would have imaginary solutions if b² - 4ac is negative. That is:

b² - 4ac < 0

Given an equation: 3x²+2kx+2=0

a) If k = -5

a= 3, b = 2k = 2 × -5 = -10, c= 2

b² - 4ac = (-10)² - 4(3)(2) = 76 > 0

It would not produce imaginary solutions

b) If k = -3

a= 3, b = 2k = 2 × -3 = -6, c= 2

b² - 4ac = (-6)² - 4(3)(2) = 12 > 0

It would not produce imaginary solutions

c) If k = -1

a= 3, b = 2k = 2 × -1 = -2, c= 2

b² - 4ac = (-1)² - 4(3)(2) = -23 < 0

It would produce imaginary solutions

d) If k = 2

a= 3, b = 2k = 2 × 2 = 4, c= 2

b² - 4ac = (4)² - 4(3)(2) = -8 <0

It would  produce imaginary solutions

e) If k = 3

a= 3, b = 2k = 2 × 3 = 6, c= 2

b² - 4ac = (6)² - 4(3)(2) = 12 > 0

It would not produce imaginary solutions

f) If k = -5

a= 3, b = 2k = 2 × 4 = 8, c= 2

b² - 4ac = (8)² - 4(3)(2) = 40 > 0

It would not produce imaginary solutions

2)

A. Distribute 4 to the quantity ( x+1). to get:

1/2 x+12=8+4x+4

B. Add − 8 to each side of the equation. to get:

1/2 x+4=4( x+1)

C. Distribute 12 to the quantity ( x+1). This is not correct

D. Multiply each side of the equation by 2. to get:

x + 24 = 16 + 8(x +1)

F. Add together the 8 and the 4 on the right side of the equation. This is not correct

G. Multiply the left side of the equation by 2 to eliminate the fraction. This is not correct

3)

[tex]f(x)=x^2-2(r-s)x+(r-s)^2+(r+s)\\f(x)=x^2-2rx+2sx+r^2+s^2-2rs+r+s\\f(x)=x^2+r^2+s^2-2rs-2rx+2sx+r+s=(r-x-s)^2+(r+s)\\f(x)=(r-x-s)^2+(r+s)[/tex]

4) P( x )=50,000 ( 0.95 )^x, where x represents the number of years after the year 2000

At the year 2000, x = 0 therefore

[tex]P(x)=50000(0.95)^x=50000(0.95)^0=50000[/tex]

Since 0.95 is less than 1, it represent a population decay. 0.95 is the common ratio

The population is decreases by 0.95 each year

Otras preguntas