Angle(A) = ½( arc(BC) - arc(BE) )
50 = ½( (4x + 20) - (2x - 10) )
50 = ½( 4x + 20 - 2x + 10 )
50 = ½( 2x + 30 )
50 = x + 15
x = 35
Can you finish this problem? ...
let hotdog be h and bag of chip be c
Then:
\(5h+6c=18.39\) (1)
\(h+2c=4.63\) (2)
System of two equations.
(2) multiply by -3 and add to (1)
\(2h=4.5\)
\(h=2.25\)
I think that it is the sqrt(39) ...
"Simplest radical form" means to take an answer like sqrt(40) and rewrite is as 2·sqrt(10).
7.21111.... is the common fraction 649 / 90 ...
What was the problem?
hmm... That is what I thought but, I am confused that the median of AM would be a radical? I thought that the answer was 4? Srry...
Wait... would the answer be \(\sqrt{39}\)?
The median of an isosceles triangle is perpendicular to the opposite side and goes to the middle of the opposite side.
For this problem, call the midpoint of the side opposite angle(A) "M".
BM = 5 and AB = 8 and triangle ABM is a right triangle ---> AM2 + 52 = 82
Can you take it from here?
Since the triangle is an equilateral triangle, the incenter will be the center of the sphere (label this point C).
The incenter is two-thirds of the distance from a vertex to the midpoint of the opposite side; therefore LO = (3/2)·LC.
Since LC = 2m; LO = 3m
Triangle LOK is a 30-60-90 right triangle, making KO = sqrt(3)·m and LK = 2·sqrt(3)·m.
Formula for the volume of a cone: V = (1/3)·pi·r2·h = (1/3)·pi·[sqrt(3)·m]2·3m
Formula for the surface area of a cone: SA = pi·r·L + 2·pi·r2 = pi·sqrt(3)·m·2·sqrt(3)·m + 2·pi·[sqrt(3)·m]2
I'm getting "4, 2, 2".
its not saying the amont of hours
FV=500000; N=10; R=0.05;PV=FV/(1 + R)^N; print"PV =$",PV
PV =$ 306,956.63
Never mind, I solved it.
There are 20 cards.
How many are multiples of 3? 3, 6, 9, 12, 15, 18 = 6
How many are greater than 10? 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 = 10
Without counting repeats there are 13 that work.
Probability: 13/20
This is a bowl shaped parabola ( because the t^2 coefficint is POSITIVE )
the minumum 't' will be at - b/2a a =1 b = -9 c = -36
The volume of tetrahedron OABC is 70.
There must be 2 non-athletes who are left-handed. Hence probability of randomly selecting student who is athlete or left-handed (or both) is (35+2)/50 = 0.74
"Given that a and b are the roots of the equation 2x²+x+14=0 find the equation whose roots are a^4 and b^4."
(Note: I actually substituted in the order (x - b4)(x - a4))
All ideologies have failings. All deities have failings too. That’s why we need apologists.
i thought only failed ideologies have apologists but anything goes
Wow! That’s wonderful! I hope you didn’t break your finger. You now qualify for entry into paradise.
I can hear the deity now: “Welcome my good and faithful servant, you have pleased me well! In My Holy Presence you will forever dwell...”
a=46656; c=(1); n=2; s=2#a;cycle:d=a/n^2;if(a%n^2==0, c=sort(c,n),0);n++;if(n<=s, goto cycle, c);printc,;print">>Total P^2 =", count c;print;print;
OUTPUT: (1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216)>>Total P^2 = 16 factors.
Find the value of x.
I already said thank you, and I gave Geno a point.
360 degrees / 60 min = 6 degrees/min
7 min x 6 degrees / min = 42 degrees
12 min x 6 degrees/ min =......... degrees
162o / 6o/min = ........ min
