I count 17 such numbers: 2^2, 3^2, 3^3, 4^2, 5^2, 7^2, 8^2, 9^2, 11^2, 13^2, 17^2, 19^2, 23^2, 25^2, 27^2, 29^2, 31^2.
Note: Of course, I did not include any of the 168 primes, since all of them have 2 divisors, which of course is a prime number.
Close enough I guess.
I thnk that you might mean this????
tan-1 ( u) = θ means that tan (θ) = u = u / 1 = y / x
So....by the Pythagorean Theorem.....
x^2 + y^2 = r^2
1^2 + u^2 = r^2 take both roots
±√[ 1 + u^2 ] = r
So cos θ = x / r = ±1 / √[ 1 + u^2 ]
[ I forgot to take both roots in my first answer ]
Is there a way to do it using like pythagorean identities and stuff like that? That's the way my professor wants us to do it.
tan-1 (u) = θ which implies that
tan (θ) = u = u / 1
So.....tan = y / x which implies that y = u and x = 1
And cos ( θ) = x / r
So r = √ [ x^2 + y^2 ] = √[ 1^2 + u^2 ] = √ [ 1 + u^2 ]
So....putting all this together
cos ( tan-1 (u) ) = cos (θ) = x / r = 1 / √ [ 1 + u^2 ]
THX, EP....
Try this
C(9,1) = 9
So we have
9 (1/6) = 9/6 = 3/2 = 1.5
So....we really have
1.5 * (5/6)^8 ≈ .3488
.348852 I do not know why you get 9...sorry
I tried to input that into my calculator to get a decimal in the fourth place as required by my answer, but I get 9, why is that?
P ( one six) = C(9,1) ( 1/6) ( 5/6)^8 ≈ .349 = 34.9%
a) When the ball hits the ground, y will = 0
set the second equation = 0 and solve for 'hang time'
b) Max height will be = - b/2a from the second equation -16t^2 + v sin (theta) t + h0 b= v sine theta a = -16
c) use t from question a) and substitute into the x equation to find the distance
Need more help?
\(0 < a < \dfrac \pi 2 \Rightarrow 0 < \sin(a) < 1,~0 < \cos(a) < 1\\ 0<\sin(a)<1 \Rightarrow \sin^k(a) < \sin^j(a),~k>j\geq 1\\ 0<\cos(a)<1 \Rightarrow \cos^k(a) < \cos^j(a),~k>j\geq 1\)
\(\sin^7(a) < \sin^2(a)\\ \cos^7(a) < \cos^2(a)\\ \sin^7(a) + \cos^7(a) < \sin^2(a) + \cos^2(a) = 1\)
Look at the last two fractions
The sum of these = [ √2 - 2] + [ √2 + 2 ] 2√2 2√2
________________ = _____ = ___ = - √2
[ √2+ 2 ] [√2 - 2 ] 2 - 4 -2
So.....we just have
2 + √2 - √2 =
2
The curve passes through (0, -13) so -13 = a*02+b*0+c or c = -13
Hence y = ax2 + bx -13
Maximum occurs when y = 5 and x = 3, so 5 = 9a + 3b -13
or 9a + 3b = 18, or 3a + b = 6 ...(1)
The slope is given by slope = 2ax + b, and this must be zero at a maximum,
so 2a*3 + b = 0 or 6a + b = 0 ...(2)
You can use equations (1) and (2) to find a and b, then use
m = a*42 + 4b - 13 to calculate m.
Changing only the 'c' shifts the parabola up (or down) and affects all three parameters: h intercept, max h and t intercept.
We have 2^3 outcomes = 8 outcomes
There is 1 flip where there are no heads and 3 flips where there is only one head
Every other outcome will have two or more heads
So 8 - 3 - 1 = 4 ways
y = x^2 + a
y = ax
Set these equal
x^2 + a = ax
x^2 - ax + a = 0
If they intersect once.....then there is only one root which means that the discriminant = 0
a^2 - 4a = 0 factor
a( a - 4) = 0
a = 0 or
a = 4
So....the sum is 4
ok so from 1-100 you find the closest number you multiply by to get the closest to 1000. In this case it is 111. so then you know 111 numbers are divisible by 9. 1000-111=889.
889 is the answer
Ok, so first you see is it ligma? or not
Thanks!
Thank you!
\( \dfrac{x}{x+4} = -\dfrac{9}{x+3}\)
Cross- multiply
x(x + 3) = - 9 ( x + 4)
x^2 + 3x = - 9x - 36 rearrange as
x^2 + 12x + 36 = 0 factor as
(x + 6)^2 = 0
This implies that x must be - 6
Sorry that your post was flagged......I don't know why.....
Answer here : https://web2.0calc.com/questions/help_81197
