Answer: 8(8+x)=80
Step-by-step explanation: 8 times is shown in parenthesis, 8+x is the sum, and 80 is the answer
Answer:
D. $120 for 20 lessons
Step-by-step explanation:
Based on the data provided we can decide which package is best by first calculating how much each package is charging per lesson. We do this by dividing the total dollar amount of each package by the number of lessons offered in that package. Then we simply compare each package to see which package charges the least amount for every single lesson like so...
$90 / 10 = $9 per lesson
$160 / 20 = $8 per lesson
$80 / 10 = $8 per lesson
$120 / 20 = $6 per lesson
Based on these values we can see that the cheapest package is $120 for 20 lessons which would be $6 per lesson.
Answer:
D
Step-by-step explanation:
all of the numbers in the category of X cam make up the numbers in the category of Y, They represent the same ratio because you can get to 8 by counting my 2's and u can get to 12 by counting by 3's...idk if that made any sense but I hope this helps somehow.
Answer:
1.99 L
Step-by-step explanation:
In the question it is mentioned that
The total liters of the water filled is 2 liters
But the family drank 11.73 liters
How should it be possible that the family drank is more than the total liters
So here we assume that the family drunk water should be in milliliters
Now the number of liters left is
= 2,000 ML - 11.73 ML
= 1988.27 ML
Now convert into liters which is
= 1988.27 ÷ 1,000
= 1.99 L
Very nice to have an accompanied image!Illumination is proportional to the intensity of the source, inversely proportional to the distance squared, and to the sine of angle alpha.so that we can writeI(h)=K*sin(alpha)/s^2 ................(0)where K is a constant proportional to the light source, and a function of other factors.
Also, radius of the table is 4'/2=2', therefore, using Pythagoras theorem,s^2=h^2+2^2 ...........(1), and consequently,sin(alpha)=h/s=h/sqrt(h^2+2^2)..............(2)
Substitute (1) and (2) in (0), we can writeI(h)=K*(h/sqrt(h^2+4))/(h^2+4)=Kh/(h^2+4)^(3/2)
To get a maximum value of I, we equate the derivative of I (wrt alpha) to 0, orI'(h)=0or, after a few algebraic manipulations, I'(h)=K/(h^2+4)^(3/2)-(3*h^2*K)/(h^2+4)^(5/2)=K*sqrt(h^2+4)(2h^2-4)/(h^2+4)^3We see that I'(h)=0 if 2h^2-4=0, giving h=sqrt(4/2)=sqrt(2) feet above the table.
We know that I(h) is a minimum if h=0 (flat on the table) or h=infinity (very, very far away), so instinctively h=sqrt(2) must be a maximum.Mathematically, we can derive I'(h) to get I"(h) and check that I"(sqrt(2)) is negative (for a maximum). If you wish, you could go ahead and find that I"(h)=(sqrt(h^2+4)*(6*h^3-36*h))/(h^2+4)^4, and find that the numerator equals -83.1K which is negative (denominator is always positive).
An alternative to showing that it is a maximum is to check the value of I(h) in the vicinity of h=sqrt(2), say I(sqrt(2) +/- 0.01)we findI(sqrt(2)-0.01)=0.0962218KI(sqrt(2)) =0.0962250K (maximum)I(sqrt(2)+0.01)=0.0962218KIt is not mathematically rigorous, but it is reassuring, without all the tedious work.
