60.
The question gives you the fact that 6 bricks is 2 feet high.
What do you do to get from 2 to 20?
You multiply by 10.
Since you multiplied one thing by 10, you must do it to the other side so it remains balanced.
2 ft x 10ft = 20ft
6 bricks x 10 bricks = 60 bricks
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
Using the concept of domain, the domain of (f.g)(x) is given by:
{x ∈ ℝ | x ≠ 3}
---------------------
- The domain of a function is given by all possible input values, that is, <u>on a graph, all values that the x-axis assumes.</u>
- In the graph, <u>function f assumes all real values.</u>
- Function g is not defined for x = 3, thus, it's domain is all real values except 3.
- Thus, the multiplication, as
, will also not be defined at x = 3, and the domain of the multiplication is:
{x ∈ ℝ | x ≠ 3}
A similar problem is given at brainly.com/question/4175434
Answer:
SA=571.77
Step-by-step explanation:
SA==2πrh+2πr2=2·π·7·6+2·π·72≈571.76986
pls mark brainliest
