Answer:
10x<9x>=-3
Step-by-step explanation:
6x +4x <-3 (5x-8>=-3
10x<-3(-3x>=-3
multiple the bracket by -3
10x<9x>-3
Answer:
(g◦f)(x) = 2x +1
Step-by-step explanation:
(g◦f)(x) = g(f(x)) = g(2x -1) = (2x -1) +2
(g◦f)(x) = 2x +1
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Substitute f(x) for the argument in the function g(x) and simplify.
Answer:
33(5/6) sq miles
Step-by-step explanation:
Area = LxW
812 = 24xW
Divide by 24
W = 33.8333333 ( 33 (5/6)
8 is non repeating
3 is repeating
For nominator subtract the non repeating (83-8 = 75)
For denominator : one 9 for each non repeating,one zero for each repeating (only 3 is repeating,then one zero)
75/90 = 5/6
Answer:
conditions h, j, and k will form an integer
Step-by-step explanation:
any 2 integers added together will form an integer, it doesn't matter if the integers are odd or even
f and g can't be always true, since we don't know if the other number is has decimals
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
