Answer:
166⅓ %
Step-by-step explanation:
250/150 × 100
500/3 = 166⅓ %
Or 166.67%
Answer:
f(x) = 3x² + 6x + 1
General Formulas and Concepts:
- Order of Operations: BPEMDAS
- Expand by FOIL (First Outside Inside Last)
- Standard Form: f(x) = ax² + bx + c
- Vertex Form: f(x) = a(bx + c)² + d
Step-by-step explanation:
<u>Step 1: Define function</u>
f(x) = 3(x + 1)² - 2
<u>Step 2: Find Standard Form</u>
- Expand by FOILing: f(x) = 3(x² + 2x + 1) - 2
- Distribute 3: f(x) = 3x² + 6x + 3 - 2
- Combine like terms (constants): f(x) = 3x² + 6x + 1
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
Answer go to ask quastion
Step-by-step explanation:
go to ask quaiatsom
Answer:
El resultado es:
2
x
^2 − 8
x
Step-by-step explanation:
¡Espero que esto ayude! ¡Que tengas un gran día!
