Answer:
(0,-2)
Step-by-step explanation:
Answer:
∠AEI ≅ ∠DEH because vertical angles are congruent; rotate ΔHED 180° around point E, then translate point D to point A to confirm ∠IAE ≅ ∠HDE.
Step-by-step explanation:
tbh im not suuper sure but my educated guess is that by looking at it. Good Luck!
The only ones that would work are if the factor is composite and if 1 is added to it it can be divided into 48. So 15 would work. The factor 15 is composite, and if 1 is added it can be divided into 48.Therefore the son is 15.
Answer:
The answer is below
Step-by-step explanation:
The volume of a cuboid is the product of its length, height and breadth. It is given by:
Volume = length × breadth × height
Since the volume is given by the expression 12y² + 8y - 20. That is:
Volume = 12y² + 8y - 20 = 4(3y² + 2y - 5) = 4(3y² + 5y - 3y -5) = 4[y(3y + 5) -1(3y + 5)]
Volume = 4(y-1)(3y+5)
Or
Volume = 12y² + 8y - 20 = 2(6y² +4y - 10) = 2(6y² + 10y - 6y -10) = 2[y(6y + 10) -1(6y + 10)]
Volume = 2(y-1)(6y+10)
Therefore the dimensions of the cuboid are either 4, y-1 and 3y+5 or 2, y-1 and 6y+10
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