Answer:1,200
Step-by-step explanation:
Answer:
x=4/5
Step-by-step explanation:
6^x^10 = 6^12 / 6^4
We know that a^b^c = a^(b*c)
6^x^10 = 6^10x
and
We know a^b / a^c = a^(b-c)
6^12 / 6^4 = 6^(12-4) = 6^8
Replacing into the original equation
6^x^10 = 6^12 / 6^4
6^(10x) = 6^8
The bases are the same, so the exponents must be the same
10x = 8
Divide each side by 10
10x/10 = 8/10
x = 4/5
Answer:
A mushroom
Step-by-step explanation:
Answer:
2cm < BC < 12cm
Step-by-step explanation:
Answer:
∫∫∫1 dV=4\sqrt{3}π
Step-by-step explanation:
From Exercise we have
z=6-x^{2}-y^{2}
z=x^{2}+y^{2}
we get
2z=6
z=3
x^{2}+y^{2}=3
We use the polar coordinates, we get
x=r cosθ
y=r sinθ
x^{2}+y^{2}&=r^{2}
r^{2}=3
We get at the limits of the variables that well need for our integral
x^{2}+y^{2}≤z≤3
0≤r ≤\sqrt{3}
0≤θ≤2π
Therefore, we get a triple integral
\int \int \int 1\, dV&=\int \int \left(\int_{x^2+y^2}^{3} 1\, dz\right) dA
=\int \int \left(z|_{x^2+y^2}^{3} \right) dA
=\int \int\ \left(3-(x^2+y^2) \right) dA
=\int \int\ \left(3-r^2 \right) dA
=\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} (3-r^2) dr dθ
=3\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} 1 dr dθ-\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} r^2 dr dθ
=3\int_{0}^{2\pi} r|_{0}^{\sqrt{3}} dθ-\int_{0}^{2\pi} \frac{r^3}{3}|_{0}^{\sqrt{3}}dθ
=3\sqrt{3}\int_{0}^{2\pi} 1 dθ-\sqrt{3}\int_{0}^{2\pi} 1 dθ
=3\sqrt{3} ·2π-\sqrt{3}·2π
=4\sqrt{3}π
We get
∫∫∫1 dV=4\sqrt{3}π
