Answer:
275
Step-by-step explanation:
1000-725=275
Answer:
the answer is 230 calories :)
Step-by-step explanation:
divide 160 by 16 to find out how much 1-oz is then times it by 23 to get the amount of calories in 23-oz. There you go I hope it helps :)
Answer:
The height of the equilateral triangle is 24 units
Step-by-step explanation:
In the question above, the unit of the side of the equilateral triangle is not given. If the complete question is:
Triangle MNO is an equilateral triangle with sides measuring 16 square root of 3 units, then the length of each side is 16√3 units.
To find the height of the triangle, we use Pythagoras theorem
MP^2 + PO^2 = MO^2
Let MP^2 be x
x^2 + {(16√3)/2}^2 = (16√3)^2
X^2 + (256*3)/4 = 256*3
4x^2 + 768 = 768* 4
4x^2 + 768 = 3072
4x^2 = 3072-768
4x^2 = 2304
X^2 = 2304/4
X^2 = 576
x= √576
x = 24 units
Answer:
- Library 2 charges more for each book loaned.
- Library 1 has a cheaper subscription fee.
Step-by-step explanation:
Based on the table, we can write the equation for the cost of borrowing from Library 2 using the two-point form of the equation of a line:
y = (y2 -y1)/(x2 -x1)(x -x1) +y1
for (x1, y1) = (2, 15.50) and (x2, y2) = (8, 26) this equation becomes ...
y = (26 -15.50)/(8 -2)(x -2) +15.50 . . . . . fill in the values
y = (10.50/6)(x -2) +15.50 . . . . . . . . . . . . simplify a bit
y = 1.75x -3.50 +15.50 . . . . . . simplify more
In the above, we have x = number of books; y = cost. We can use "n" and "C" for those, respectively, as in the equation for Library 1. Then the monthly cost for Library 2 is ...
C = 12 + 1.75n . . . . . . . arranged to the same form as for Library 1
_____
Now, we can answer the questions.
Library 2 charges more for each book loaned. (1.75 vs 1.50 for Library 1)
Library 1 has a cheaper subscription fee. (10 vs 12 for Library 2)
_____
The numbers in the cost equations are ...
C = (subscription fee) + (cost per book loaned)·n
If the signs of the rational numbers are the same, you add them, i.e. ++ or --.
If they are different, you subtract them, i.e. +- or -+
