Answer:
Step-by-step explanation:
How do we find the line that's perpendicular to another?
We find the negative reciprocal. So for example if our slope is 2, the negative reciprocal would be .
Answer:
$30.72
Step-by-step explanation:
You were right with your answer, the teacher is probably just wanting you to tip on the subtotal (before tax) instead of the total total. Instead of doing total+tax+tip, we're going to figure out the tax and tip separately and then add it all together.
<u>Tipping on the subtotal (what your teacher wanted)</u>
20% of $24 = $4.80
8% of $24 = $1.92
$24+$4.80+$1.92=$30.72
<u>Tipping on the total (what you did)</u>
8% of $24 is $1.92
New total is $25.92
20% of $25.92 is $5.18
$25.92 + $5.18 = $31.10
First thing you should do is reduce coefficients.
1st equation has all multiples of '2'. Divide by 2
---> x +3y = -6
2nd equation has multiples of 5. Divide by 5.
---> x - y = 2
Now elimination part is easier.
Eliminate 'x' variable by subtracting 2nd equation from 1st.
x + 3y = -6
-(x - y = 2)
----------------------
4y = -8
Solve for 'y'
4y = -8
y = (-8)/4 = -2
Substitute value for 'y' back into 2nd equation:
x - (-2) = 2
x + 2 = 2
x = 0
Solution to system is:
x=0, y =-2
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.
