Answer:
−4<x≤9
Conjunction
Step-by-step explanation:
-7 < x -3
x -3 ≤ 6
-7 + 3 < ( x - 3) + 3
(x -3) + ≤ 6 + 3
-4 < x -3 + 3
x - 3 + 3 ≤ 9
-4 < x
x ≤ 9
x > -4
x ≤ 9
No it is not possible if the pen has a rectangular or square shape
The mean of 5 and 15 is 10. (5+15= 20 20/2=10)
Compare 1/7 to consecutive multiples of 1/9. This is easily done by converting the fractions to a common denominator of LCM(7, 9) = 63:
1/9 = 7/63
2/9 = 14/63
while
1/7 = 9/63
Then 1/7 falls between 1/9 and 2/9, so 1/7 = 1/9 plus some remainder. In particular,
1/7 = 1/9¹ + 2/63.
We do the same sort of comparison with the remainder 2/63 and multiples of 1/9² = 1/81. We have LCM(63, 9²) = 567, and
1/9² = 7/567
2/9² = 14/567
3/9² = 21/567
while
2/63 = 18/567
Then
2/63 = 2/9² + 4/567
so
1/7 = 1/9¹ + 2/9² + 4/567
Compare 4/567 with multiples of 1/9³ = 1/729. LCM(567, 9³) = 5103, and
1/9³ = 7/5103
2/9³ = 14/5103
3/9³ = 21/5103
4/9³ = 28/5103
5/9³ = 35/5103
6/9³ = 42/5103
while
4/567 = 36/5103
so that
4/567 = 5/9³ + 1/5103
and so
1/7 = 1/9¹ + 2/9² + 5/9³ + 1/5103
Next, LCM(5103, 9⁴) = 45927, and
1/9⁴ = 7/45927
2/9⁴ = 14/45927
while
1/5103 = 9/45927
Then
1/5103 = 1/9⁴ + 2/45927
so
1/7 = 1/9¹ + 2/9² + 5/9³ + 1/9⁴ + 2/45927
One last time: LCM(45927, 9⁵) = 413343, and
1/9⁵ = 7/413343
2/9⁵ = 14/413343
3/9⁵ = 21/413343
while
2/45927 = 18/413343
Then
2/45927 = 2/9⁵ + remainder
so
1/7 = 1/9¹ + 2/9² + 5/9³ + 1/9⁴ + 2/9⁵ + remainder
Then the base 9 expansion of 1/7 is
0.12512..._9
Answer:
y= -(1/3)x -r
Step-by-step explanation:
a perpendicular line has the slope of the line that it's perpendicular but its the negative reciprocal of it. So for ex: line 1 = 2x then the line perpendicular to it is -1/2x.
In this case, the slope is 3x. The line perpendicular to it therefore, should be -1/3x.
The problem states the line has the same y-intercept so it has to still be -r.
So the equation of the line perpendicular to y=3x-r with the same y-intercept is y=-(1/3)x -r
