Third term = t3 = ar^2 = 444 eq. (1)
Seventh term = t7 = ar^6 = 7104 eq. (2)
By solving (1) and (2) we get,
ar^2 = 444
=> a = 444 / r^2 eq. (3)
And ar^6 = 7104
(444/r^2)r^6 = 7104
444 r^4 = 7104
r^4 = 7104/444
= 16
r2 = 4
r = 2
Substitute r value in (3)
a = 444 / r^2
= 444 / 2^2
= 444 / 4
= 111
Therefore a = 111 and r = 2
Therefore t6 = ar^5
= 111(2)^5
= 111(32)
= 3552.
<span>Therefore the 6th term in the geometric series is 3552.</span>
Answer: 19.2inches³
Step-by-step explanation:
The volume of a cone is calculated as:
= 1/3πr²h
where,
π = 3.14
r = radius = diameter/2 = 3.5/2 = 1.75
h = height = 6
Volume = 1/3πr²h
= 1/3 × 3.14 × 1.75² × 6
= 19.2325
= 19.2inches³
Therefore, the volume of the cone-shaped sandwich is 19.2inches³.
Answer:
Step-by-step explanation:
Make denominators same:
Add the numerators:
Apply the rule:
Therefore,
Cancel out common factor of 2x+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
