First one
when doing the quadratic equation, we come across something called the determinant

when this is positive, we have 2 real roots
when this is 0, we have 1 real root
whenthis is negative, we have 2 imaginary rots
we want 2imaginary roots so

<0
we have
1x^2-6x+c=0
a=1
b=-6
c=c

<0

<0
square root both sies
36-4c<0
add 4c both sides
36<4c
divide both sides by 4
9<c
c>9
D is the answer (not b or c)
remember
√-1=i
a²-b²=(a-b)(a+b)
we want to do
x^2-(-9)
now we have to take the sqrt of -9
√-9=(√-1)(√9)=(i)(3)=3i
(x)^3-(3i)^2
(x-3i)(x+3i)
B is answer
28 + 8*10.5 = 28 + 84 = 112
The weekly cost is $112
Answer:
y=2x+6 ; x+y=51
Step-by-step explanation:
assuming x is one number and y is the other -
one number (y) is 6 more (+6) than twice another (2x) -
using that knowledge we form an equation -
y=2x+6
the sum (x+y) is 51
using this knowledge we can make the equation -
x+y=51
now you have -
y=2x+6 or 2x-y=-6
x+y=51
using system of equations -
you add the equations -
3x=45
x=15
y=51-x => y=36
hope this helps!!
Answer:
y = √(3² + 3²)
9 = √(5² + x²)
Step-by-step explanation:
Diagram 1:
Hypotenuse = y
Opposite = 3
Adjacent = 3
y² = 3² + 3²
y = √(3² + 3²)
y = √(9 + 9)
y = √18
Diagram 2:
Hypotenuse ² = known leg² + unknown leg²
Hypotenuse = 9
Known leg = 5
Unknown leg = x
9² = 5² + x²
9 = √(5² + x²)
9² = 5² + x²
9² - 5² = x²
x² = 81 - 25
x = √ 56