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find the volume of the solid. round your answer to two decimal places. …

Question

find the volume of the solid. round your answer to two decimal places.
1.
2.
3.
4.
5.
6.

Explanation:

  1. For the cone in question 1:
  • The volume formula for a cone is $V=\frac{1}{3}\pi r^{2}h$. Here, $r = 4$ cm and $h=6$ cm.
  • # Explanation:
  • ## Step1: Calculate the base - area
  • The base - area of a cone (a circle) is $A=\pi r^{2}$. Substituting $r = 4$ cm, we get $A=\pi\times4^{2}=16\pi$ $cm^{2}$.
  • ## Step2: Calculate the volume
  • Using the volume formula $V=\frac{1}{3}Ah$, and substituting $A = 16\pi$ $cm^{2}$ and $h = 6$ cm, we have $V=\frac{1}{3}\times16\pi\times6=32\pi\approx100.53$ $cm^{3}$.
  • # Answer:
  • $100.53$ $cm^{3}$
  1. For the rectangular - based pyramid in question 2:
  • The volume formula for a rectangular - based pyramid is $V=\frac{1}{3}lwh$, where $l = 5$ in, $w = 3$ in and $h = 4$ in.
  • # Explanation:
  • ## Step1: Calculate the base - area
  • The base - area of the rectangular - based pyramid is $A=lw$. Substituting $l = 5$ in and $w = 3$ in, we get $A=5\times3 = 15$ $in^{2}$.
  • ## Step2: Calculate the volume
  • Using the volume formula $V=\frac{1}{3}Ah$, and substituting $A = 15$ $in^{2}$ and $h = 4$ in, we have $V=\frac{1}{3}\times15\times4 = 20$ $in^{3}$.
  • # Answer:
  • $20$ $in^{3}$
  1. For the triangular - based pyramid in question 3:
  • First, find the area of the equilateral - triangle base. The area formula for an equilateral triangle with side length $s$ is $A=\frac{\sqrt{3}}{4}s^{2}$. Here, $s = 4$ cm, so $A=\frac{\sqrt{3}}{4}\times4^{2}=4\sqrt{3}$ $cm^{2}$. The volume formula for a pyramid is $V=\frac{1}{3}Ah$, with $h = 4$ cm.
  • # Explanation:
  • ## Step1: Calculate the base - area
  • Using the formula for the area of an equilateral triangle $A=\frac{\sqrt{3}}{4}s^{2}$ with $s = 4$ cm, we get $A = 4\sqrt{3}$ $cm^{2}$.
  • ## Step2: Calculate the volume
  • Using the volume formula $V=\frac{1}{3}Ah$, and substituting $A = 4\sqrt{3}$ $cm^{2}$ and $h = 4$ cm, we have $V=\frac{1}{3}\times4\sqrt{3}\times4=\frac{16\sqrt{3}}{3}\approx9.24$ $cm^{3}$.
  • # Answer:
  • $9.24$ $cm^{3}$
  1. For the cone in question 4:
  • The volume formula for a cone is $V=\frac{1}{3}\pi r^{2}h$. Here, $r = 6$ m and $h = 11$ m.
  • # Explanation:
  • ## Step1: Calculate the base - area
  • The base - area of a cone (a circle) is $A=\pi r^{2}$. Substituting $r = 6$ m, we get $A=\pi\times6^{2}=36\pi$ $m^{2}$.
  • ## Step2: Calculate the volume
  • Using the volume formula $V=\frac{1}{3}Ah$, and substituting $A = 36\pi$ $m^{2}$ and $h = 11$ m, we have $V=\frac{1}{3}\times36\pi\times11 = 132\pi\approx415.48$ $m^{3}$.
  • # Answer:
  • $415.48$ $m^{3}$
  1. For the rectangular - based pyramid in question 5:
  • The volume formula for a rectangular - based pyramid is $V=\frac{1}{3}lwh$, where $l = 7$ in, $w = 6$ in and $h = 9$ in.
  • # Explanation:
  • ## Step1: Calculate the base - area
  • The base - area of the rectangular - based pyramid is $A=lw$. Substituting $l = 7$ in and $w = 6$ in, we get $A=7\times6 = 42$ $in^{2}$.
  • ## Step2: Calculate the volume
  • Using the volume formula $V=\frac{1}{3}Ah$, and substituting $A = 42$ $in^{2}$ and $h = 9$ in, we have $V=\frac{1}{3}\times42\times9=126$ $in^{3}$.
  • # Answer:
  • $126$ $in^{3}$
  1. For the cone in question 6:
  • The volume formula for a cone is $V=\frac{1}{3}\pi r^{2}h$. Here, $r = 5$ cm and $h = 8$ cm.
  • # Explanation:
  • ## Step1: Calculate the base - area
  • The base - area of a cone (a circle) is $A=\pi r^{2}$. Substituting $r = 5$ cm, we get $A=\pi\times5^{2}=25\pi$ $cm^{2}$.
  • ## Ste…

Answer:

  1. For the cone in question 1:
  • The volume formula for a cone is $V=\frac{1}{3}\pi r^{2}h$. Here, $r = 4$ cm and $h=6$ cm.
  • # Explanation:
  • ## Step1: Calculate the base - area
  • The base - area of a cone (a circle) is $A=\pi r^{2}$. Substituting $r = 4$ cm, we get $A=\pi\times4^{2}=16\pi$ $cm^{2}$.
  • ## Step2: Calculate the volume
  • Using the volume formula $V=\frac{1}{3}Ah$, and substituting $A = 16\pi$ $cm^{2}$ and $h = 6$ cm, we have $V=\frac{1}{3}\times16\pi\times6=32\pi\approx100.53$ $cm^{3}$.
  • # Answer:
  • $100.53$ $cm^{3}$
  1. For the rectangular - based pyramid in question 2:
  • The volume formula for a rectangular - based pyramid is $V=\frac{1}{3}lwh$, where $l = 5$ in, $w = 3$ in and $h = 4$ in.
  • # Explanation:
  • ## Step1: Calculate the base - area
  • The base - area of the rectangular - based pyramid is $A=lw$. Substituting $l = 5$ in and $w = 3$ in, we get $A=5\times3 = 15$ $in^{2}$.
  • ## Step2: Calculate the volume
  • Using the volume formula $V=\frac{1}{3}Ah$, and substituting $A = 15$ $in^{2}$ and $h = 4$ in, we have $V=\frac{1}{3}\times15\times4 = 20$ $in^{3}$.
  • # Answer:
  • $20$ $in^{3}$
  1. For the triangular - based pyramid in question 3:
  • First, find the area of the equilateral - triangle base. The area formula for an equilateral triangle with side length $s$ is $A=\frac{\sqrt{3}}{4}s^{2}$. Here, $s = 4$ cm, so $A=\frac{\sqrt{3}}{4}\times4^{2}=4\sqrt{3}$ $cm^{2}$. The volume formula for a pyramid is $V=\frac{1}{3}Ah$, with $h = 4$ cm.
  • # Explanation:
  • ## Step1: Calculate the base - area
  • Using the formula for the area of an equilateral triangle $A=\frac{\sqrt{3}}{4}s^{2}$ with $s = 4$ cm, we get $A = 4\sqrt{3}$ $cm^{2}$.
  • ## Step2: Calculate the volume
  • Using the volume formula $V=\frac{1}{3}Ah$, and substituting $A = 4\sqrt{3}$ $cm^{2}$ and $h = 4$ cm, we have $V=\frac{1}{3}\times4\sqrt{3}\times4=\frac{16\sqrt{3}}{3}\approx9.24$ $cm^{3}$.
  • # Answer:
  • $9.24$ $cm^{3}$
  1. For the cone in question 4:
  • The volume formula for a cone is $V=\frac{1}{3}\pi r^{2}h$. Here, $r = 6$ m and $h = 11$ m.
  • # Explanation:
  • ## Step1: Calculate the base - area
  • The base - area of a cone (a circle) is $A=\pi r^{2}$. Substituting $r = 6$ m, we get $A=\pi\times6^{2}=36\pi$ $m^{2}$.
  • ## Step2: Calculate the volume
  • Using the volume formula $V=\frac{1}{3}Ah$, and substituting $A = 36\pi$ $m^{2}$ and $h = 11$ m, we have $V=\frac{1}{3}\times36\pi\times11 = 132\pi\approx415.48$ $m^{3}$.
  • # Answer:
  • $415.48$ $m^{3}$
  1. For the rectangular - based pyramid in question 5:
  • The volume formula for a rectangular - based pyramid is $V=\frac{1}{3}lwh$, where $l = 7$ in, $w = 6$ in and $h = 9$ in.
  • # Explanation:
  • ## Step1: Calculate the base - area
  • The base - area of the rectangular - based pyramid is $A=lw$. Substituting $l = 7$ in and $w = 6$ in, we get $A=7\times6 = 42$ $in^{2}$.
  • ## Step2: Calculate the volume
  • Using the volume formula $V=\frac{1}{3}Ah$, and substituting $A = 42$ $in^{2}$ and $h = 9$ in, we have $V=\frac{1}{3}\times42\times9=126$ $in^{3}$.
  • # Answer:
  • $126$ $in^{3}$
  1. For the cone in question 6:
  • The volume formula for a cone is $V=\frac{1}{3}\pi r^{2}h$. Here, $r = 5$ cm and $h = 8$ cm.
  • # Explanation:
  • ## Step1: Calculate the base - area
  • The base - area of a cone (a circle) is $A=\pi r^{2}$. Substituting $r = 5$ cm, we get $A=\pi\times5^{2}=25\pi$ $cm^{2}$.
  • ## Step2: Calculate the volume
  • Using the volume formula $V=\frac{1}{3}Ah$, and substituting $A = 25\pi$ $cm^{2}$ and $h = 8$ cm, we have $V=\frac{1}{3}\times25\pi\times8=\frac{200\pi}{3}\approx209.44$ $cm^{3}$.
  • # Answer:
  • $209.44$ $cm^{3}$