A pharmacist has 40% and 80% of iodine solutions on hand. How many liters of each iodine solution will be required to produce 4 liters of a 50% iodine mixture?
.
Let x = liters of 40% iodine
then
4-x = liters of 80% iodine
Using algebra:
.40x + .80(4-x) = .50(4)
.40x + 3.20-.80x = 2
3.20-.40x = 2
x = 4 liters (40% iodine)
80% iodine:
4-x = 4-4 = 0 liters needed (80% iodine)
Answer:
47x - 9
Step-by-step explanation:
-4(4x + 4) + 7(1 + 9x)
-16x -16 + 7 + 63x
47x -9
You can find what x = because there is no = sign. This is in simplest form.
Answer: The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2 = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.
Step-by-step explanation: this is the same paragraph The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2 = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.
Parallel lines have equal slopes. so the equation y=-4x+5.has slope of -4 because this is in slope intercept form y=mx+b where m is slope.
so we know a parallel line will have slope -4
we have a point on the line so we will use point slope formula to solve.
y-y1=m(x-x1) where (x1,y1) is point on line
so we get
y-4=-4(x-0) to put in slope intercept form we solve for y
y-4=-4x
y=-4x+4 is an equation of a line parallel to y=-4+5 passing through (0,4)
