Answer:
Step-by-step explanation:
Where are the choices?
m and f are the cups of milk and flour, respectively.
m = ¼f + 1
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work
EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same
For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y
For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4
Therefore, for the two expressions to be conjugates, we must satisfy the two conditions.
Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the
x²y = -4 ... (I)
Condition 2: Real parts are the same
x² + y = -3 ... (II)
We have a system of equations since both conditions must be satisfied
x²y = -4 ... (I)
x² + y = -3 ... (II)
We can rearrange equation (II) so that we have
y = -3 - x² ... (II)
Substituting into equation (I)
x²y = -4 ... (I)
x²(-3 - x²) = -4
-3x² - x⁴ = -4
x⁴ + 3x² - 4 = 0
(x² + 4)(x² - 1) = 0
(x² + 4)(x-1)(x+1) = 0
Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.
Solve for y:
y = -3 - x² ... (II)
y = -3 - (±1)²
y = -3 - 1
y = -4
So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:
-3 + ix²y
= -3 + i(±1)²(-4)
= -3 - 4i
x² + y + 4i
= (±1)² - 4 + 4i
= 1 - 4 + 4i
= -3 + 4i
They result in conjugates
Answer:
A and C
Step-by-step explanation:
To determine which events are equal, we explicitly define the elements in each set builder.
For event A
A={1.3}
for event B
B={x|x is a number on a die}
The possible numbers on a die are 1,2,3,4,5 and 6. Hence event B is computed as
B={1,2,3,4,5,6}
for event C
![C=[x|x^{2}-4x+3]\\solving x^{2}-4x+3\\x^{2}-4x+3=0\\x^{2}-3x-x+3=0\\x(x-3)-1(x-3)=0\\x=3 or x=1](https://tex.z-dn.net/?f=C%3D%5Bx%7Cx%5E%7B2%7D-4x%2B3%5D%5C%5Csolving%20%20x%5E%7B2%7D-4x%2B3%5C%5Cx%5E%7B2%7D-4x%2B3%3D0%5C%5Cx%5E%7B2%7D-3x-x%2B3%3D0%5C%5Cx%28x-3%29-1%28x-3%29%3D0%5C%5Cx%3D3%20or%20x%3D1)
Hence the set c is C={1,3}
and for the set D {x| x is the number of heads when six coins re tossed }
In the tossing a six coins it is possible not to have any head and it is possible to have head ranging from 1 to 6
Hence the set D can be expressed as
D={0,1,2,3,4,5,6}
In conclusion, when all the set are compared only set A and set C are equal
Answer:
d
Step-by-step explanation:
when f(x)=5......It can also be
f(x)=1/125(625x)
every whole number over one...1(625x)=625x
also 125(1)=125 therefore (625/125)x=5x
Answer:
The answer is 32
Step-by-step explanation:
12 plus 20
