Answer:
![105m^3/min](https://tex.z-dn.net/?f=105m%5E3%2Fmin)
Step-by-step explanation:
We are given that
Base side of square pyramid, a=3 m
Height of square pyramid, h=9m
![\frac{da}{dt}=6m/min](https://tex.z-dn.net/?f=%5Cfrac%7Bda%7D%7Bdt%7D%3D6m%2Fmin)
![\frac{dh}{dt}=-1/min](https://tex.z-dn.net/?f=%5Cfrac%7Bdh%7D%7Bdt%7D%3D-1%2Fmin)
We have to find the rate of change of the volume of the pyramid at that instant.
Volume of square pyramid, V=![\frac{1}{3}a^2h](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7Da%5E2h)
Differentiate w.r.t t
![\frac{dV}{dt}=\frac{1}{3}(2ah\frac{da}{dt}+a^2\frac{dh}{dt})](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B3%7D%282ah%5Cfrac%7Bda%7D%7Bdt%7D%2Ba%5E2%5Cfrac%7Bdh%7D%7Bdt%7D%29)
Substitute the values
![\frac{dV}{dt}=\frac{1}{3}(2(3)(9)(6)+(3^2)(-1)](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B3%7D%282%283%29%289%29%286%29%2B%283%5E2%29%28-1%29)
![\frac{dV}{dt}=105m^3/min](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D105m%5E3%2Fmin)
Hence, the rate of change of the volume of the pyramid at that instant=![105m^3/min](https://tex.z-dn.net/?f=105m%5E3%2Fmin)
10x+3 = 3x+24
10x-3x= 24-3
7x=21
X= 21/7
Final answer:
X=3
I’m not so sure of my answer but wish you the best and good luck!!
Answer:
x = 2/5
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtract Property of Equality
Step-by-step explanation:
<u>Step 1: Define</u>
5x = 2
<u>Step 2: Solve for </u><em><u>x</u></em>
- Divide 5 on both sides: x = 2/5
<u>Step 3: Check</u>
<em>Plug in x into the original equation to verify it's a solution.</em>
- Substitute in <em>x</em>: 5(2/5) = 2
- Multiply: 2 = 2
Here we see that 2 does indeed equal 2.
∴ x = 2/5 is the solution to the equation.
Answer:
1. 3
2. 12
3. 27
Step-by-step explanation: