AB = 2 - sqrt(3).
x = 47 degrees.
By the way, the answer is (-inf,-1) U (1,3) U (3,inf).
There are: 900 / 5 ==180 multiples of 5
There are: 900 / 8 ==112 multiples of 8
There are: 900 / 40 ==22 multiples of 40
900 - 180 - 112 + 22 ==630 integers that are neither multiples of 5 or 8
Thanks!
simplify (48 a^5 b^3)/(3 a^2 b) ==16 a^3 b^2
You're close! It simplifies to 16*a^3*b.
Thank you so much! I was confused and needed help. So thanks!
https://web2.0calc.com/questions/help-please_74857#r2
The smallest multiple of 5 is \(100 \Rightarrow (20 \times 5)\), and the largest multiple of 5 is \(995 \Rightarrow (199 \times 5)\). There are \(199 - 20 + 1 = 180\) multiples of 5
The smallest multiple of 8 is \(104 \Rightarrow (13 \times 8)\), and the largest multiple of 5 is \(995 \Rightarrow (124 \times 8)\). There are \(124 - 13 + 1 = 122\) multiples of 8
However, we overcount every \(\text{lcm}(5,8) = 40\) numbers.
The smallest number we overcount for is \(120 \Rightarrow (3 \times 40)\), and the largest number is \(960 \Rightarrow (24 \times 40)\). There are \(24 - 3 + 1 = 22\) numbers we overcount for.
Can you take it from here?
\(2x-5\le \:-x+12-3x+19\)
\(2x-5\le \:-x-3x+12+19\)
\(2x-5\le \:-4x+12+19\)
\(2x-5\le \:-4x+31\)
\(2x\le \:-4x+36\)
\(6x\le \:36\)
\(x\le \:6\)
Interval notation: \((-\infty \:,\:6]\)
-Vinculum
The question is not complete
\(-4\left(x+4\right)>x+7+5\left(x+2\right)\)
\(-4x-16>6x+17\)
\(-4x>6x+33\)
\(-10x>33\)
\(10x<-33\)
\(x<-\frac{33}{10}\)
\(Interval\:Notation: \:\left(-\infty \:,\:-\frac{33}{10}\right)\)
Hi Guest,
So...
\(\frac{y}{3}+1=\frac{\left(y+3\right)}{y}\)
\(y^2+3y=3\left(y+3\right)\)
\(y=3,\:y=-3\)
The number of values of y: \(2\)
https://web2.0calc.com/questions/geometry_87697
what is the distance between each wall?
missing information
Hmm! What about y = -3?
There is only one value of y that satisfies the equation:
y ==3
[3/3+1 ==(3+3)/3]
2 == 2
(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 24, 25, 28, 30, 36, 48, 49, 50, 72, 75, 98, 144, 150, 196)==29 such divisors
1/ ( a^3 + 7) - 7 = -a^3 / ( a^3 + 7) rearrange as
1 / (a^3 + 7) + a^3/ (a^3 + 7) = 7
(1 + a^3) = 7 ( a^3 + 7)
1 + a^3 7a^3 + 49
-6a^3 = 48
a^3 = -48 / 6
a^3 = -8
a = -2
Total no. of jelly beans= 108 The ratio in which these are needed to be distributed= 4:5 So, the no. of jelly beans received by Skyler= (4*108)/9 =48 no. of jelly beans received by Tommy=(5*108)/9 =60 Hence, Skyler got 12 jelly beans lesser than Tommy
115.05
(kc + l)/(mc + n) = 3/16.
