Answer:
Step-by-step explanation:
16% is the answer.
440 is what percent of 2750?
440 = x/100 x 2750 (of means x or times in math)
440 = 2750x/100
440 = 275x/10
440x10 = 275x/10
4400 = 275x
4400/275= 275x/275
4400/275 = x
4400/275 = 16%
10)
60in=60in × (1ft/12in) = 5ft
30in= 30in × (1ft/12in) = 2.5ft
12in = 12in × (1ft/12in) = 1ft
answer is = 5ft × 2.5ft× 1ft =12.5 ft^3 = a
11)
dozen = 12 hammers
12 hammers = 20 pounds
therefore 1 hammer= 20/12 = 1.6666 lb
21 hammers ×1.666 = 35 = c
12)
convert qt to gal
600 gal × 4 qt/gal = 2400 qt
2400 qt ÷ 120 qt/min
=20 min = d
13)
Carla is 12
rob is 12+5= 17
cousin is 12-7 = 5
12+17+5= 34 =a
15)
convert 60ft into yards
60 ft × 0.333 yd/ft =20 yd
diameter = 20 yd
area of circle = (pi/4)×d^2
(pi/4)×20^2= 314.16 = b
Answer:
75 years.
Step-by-step explanation:
Divide 150 by 2, that gives you the years.
Answer:
Step-by-step explanation:
Let's say the Physics book costs $x
The Chemistry book will cost $x - $6
Two such Chemistry books will cost;
2($x - $6) = $(2x - 12)
Three such Physics books cost $3x
$(2x - 12) + $3x = $123
$5x - $12 = $123
$5x = $111
x = $22
So the Physics book costs; $22
and the Chemistry book costs $22 - 6 = $16
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
