Answer:
750 < p < 1500
Step-by-step explanation:
The total cost of the affair for p attendees is ...
750 +2.25p
The average cost is that number divided by p. The answer choices suggest that "between" means that the cost per person cannot be 2.75 or 3.25. Applying the limits to the average cost, we get ...
2.75 < (750 +2.25p)/p < 3.25
2.75 < 750/p + 2.25 < 3.25 . . . . . . . . separating the fraction parts
0.50 < 750/p < 1.00 . . . . . . . . . . . . . . subtract 2.25
2 > p/750 > 1 . . . . . . . . . . . . . . . . . . . . . take the reciprocal*
1500 > p > 750 . . . . . . . . matches choice 4
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* Taking the reciprocals of numbers with the same sign reverses their order:
2 < 3 but 1/2 > 1/3
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An alternate approach to the solution would be to multiply by p:
0.50p < 750 < 1.00p
To solve this requires it be made into two inequalities:
0.50p < 750 ⇒ p < 1500
and
750 < 1.00p ⇒ 750 < p
Then these would need to be recombined to form the answer:
750 < p < 1500
Volume is legnth times widht times height
lenght=2x-1
width=x-2
height=x+1
multiply all together
use mass distributive property
distributive=a(b+c)=ab+ac so extending that
(a+c)(c+c)=(a+b)(c)+(a+b)(d) then keep distributing so
(2x-1)(x-2)(x+1)
do each one seperately
do the first two first and put the other one (x+1) to the side for later
(2x-1)(x-2)=(2x-1)(x)+(2x-1)(-2)=(2x^2-x)+(-4x+2)=2x^2-5x+2
then do the other one
(x+1)(2x^2-5x+2)=(x)(2x^2-5x+2)+(1)(2x^2-5x+2)=(2x^3-5x^2+2x)+(2x^2-5x+2)=2x^3-3x^2-3x+2
the lasst form is 2x^3-3x^2-3x+2
Answer:
ok so use the degrees
Step-by-step explanation:
x should be 8 + 17 and then multiply the 110 degrees by 8 + 17
Answer:
0, for q ≠ 0 and q ≠ 1
Step-by-step explanation:
Assuming q ≠ 0, you want to find the value of x such that ...
q^x = 1
This is solved using logarithms.
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x·log(q) = log(1) = 0
The zero product rule tells us this will have two solutions:
x = 0
log(q) = 0 ⇒ q = 1
If q is not 0 or 1, then its value is 1 when raised to the 0 power. If q is 1, then its value will be 1 when raised to <em>any</em> power.
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<em>Additional comment</em>
The applicable rule of logarithms is ...
log(a^b) = b·log(a)
