Answer: im not completley sure but , i think its 10
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
Just took it
Answer:
<em>Proof in explanation</em>
Step-by-step explanation:
<u>Trigonometric Identities</u>
The basic trigonometric identity is:
![\sin^2\theta+\cos^2\theta=1](https://tex.z-dn.net/?f=%5Csin%5E2%5Ctheta%2B%5Ccos%5E2%5Ctheta%3D1)
We'll use it and some basic algebra to prove that, given:
![a sin^3\theta+b cos^3\theta=sin\theta cos\theta](https://tex.z-dn.net/?f=a%20sin%5E3%5Ctheta%2Bb%20cos%5E3%5Ctheta%3Dsin%5Ctheta%20cos%5Ctheta)
And
![a\sin\theta-b\cos\theta=0](https://tex.z-dn.net/?f=a%5Csin%5Ctheta-b%5Ccos%5Ctheta%3D0)
Then
![a^2+b^2=1](https://tex.z-dn.net/?f=a%5E2%2Bb%5E2%3D1)
From the equation:
![a\sin\theta-b\cos\theta=0](https://tex.z-dn.net/?f=a%5Csin%5Ctheta-b%5Ccos%5Ctheta%3D0)
We have:
![a\sin\theta=b\cos\theta\qquad [1]](https://tex.z-dn.net/?f=a%5Csin%5Ctheta%3Db%5Ccos%5Ctheta%5Cqquad%20%5B1%5D)
The equation
![a sin^3\theta+b cos^3\theta=sin\theta cos\theta](https://tex.z-dn.net/?f=a%20sin%5E3%5Ctheta%2Bb%20cos%5E3%5Ctheta%3Dsin%5Ctheta%20cos%5Ctheta)
Can be rewritten as
![a\sin\theta \sin^2\theta+b \cos^3\theta=\sin\theta \cos\theta](https://tex.z-dn.net/?f=a%5Csin%5Ctheta%20%5Csin%5E2%5Ctheta%2Bb%20%5Ccos%5E3%5Ctheta%3D%5Csin%5Ctheta%20%5Ccos%5Ctheta)
Replacing [1]:
![b\cos\theta \sin^2\theta+b \cos^3\theta=\sin\theta \cos\theta](https://tex.z-dn.net/?f=b%5Ccos%5Ctheta%20%5Csin%5E2%5Ctheta%2Bb%20%5Ccos%5E3%5Ctheta%3D%5Csin%5Ctheta%20%5Ccos%5Ctheta)
Taking the common factor:
![b\cos\theta (\sin^2\theta+ \cos^2\theta)=\sin\theta \cos\theta](https://tex.z-dn.net/?f=b%5Ccos%5Ctheta%20%28%5Csin%5E2%5Ctheta%2B%20%5Ccos%5E2%5Ctheta%29%3D%5Csin%5Ctheta%20%5Ccos%5Ctheta)
The expression in parentheses is 1, thus:
![b\cos\theta =\sin\theta \cos\theta](https://tex.z-dn.net/?f=b%5Ccos%5Ctheta%20%3D%5Csin%5Ctheta%20%5Ccos%5Ctheta)
Dividing by ![\cos\theta](https://tex.z-dn.net/?f=%5Ccos%5Ctheta)
![b=\sin\theta](https://tex.z-dn.net/?f=b%3D%5Csin%5Ctheta)
Replacing in
![a\sin\theta=b\cos\theta](https://tex.z-dn.net/?f=a%5Csin%5Ctheta%3Db%5Ccos%5Ctheta)
We have
![a\sin\theta=\sin\theta\cos\theta](https://tex.z-dn.net/?f=a%5Csin%5Ctheta%3D%5Csin%5Ctheta%5Ccos%5Ctheta)
Dividing by ![\sin\theta](https://tex.z-dn.net/?f=%5Csin%5Ctheta)
![a=\cos\theta](https://tex.z-dn.net/?f=a%3D%5Ccos%5Ctheta)
Now:
![a^2+b^2=(\cos\theta)^2+(\sin\theta)^2](https://tex.z-dn.net/?f=a%5E2%2Bb%5E2%3D%28%5Ccos%5Ctheta%29%5E2%2B%28%5Csin%5Ctheta%29%5E2)
This expression is 1, thus it's proven:
![\boxed{a^2+b^2=1}](https://tex.z-dn.net/?f=%5Cboxed%7Ba%5E2%2Bb%5E2%3D1%7D)
Answer:
Step-by-step explanation:
Volume of the pyramid :
![V=\dfrac{x^2*h}{3} =554.9\\\\x^2*h=554.9*3\\\\h=15.1\\\\x^2*15.1=554.9*3\\\\x^2=\dfrac{554.9*3}{15.1} \\\\x=\sqrt{\dfrac{1664,7}{15.1} } \\\\x=\sqrt{110,24503311258278145695364238411...} \\\\x=10,499763478887644993681148554978...\approx{10.50}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7Bx%5E2%2Ah%7D%7B3%7D%20%3D554.9%5C%5C%5C%5Cx%5E2%2Ah%3D554.9%2A3%5C%5C%5C%5Ch%3D15.1%5C%5C%5C%5Cx%5E2%2A15.1%3D554.9%2A3%5C%5C%5C%5Cx%5E2%3D%5Cdfrac%7B554.9%2A3%7D%7B15.1%7D%20%5C%5C%5C%5Cx%3D%5Csqrt%7B%5Cdfrac%7B1664%2C7%7D%7B15.1%7D%20%7D%20%5C%5C%5C%5Cx%3D%5Csqrt%7B110%2C24503311258278145695364238411...%7D%20%5C%5C%5C%5Cx%3D10%2C499763478887644993681148554978...%5Capprox%7B10.50%7D)