Anything that is behind the decimal point ends with a "ths" or "th"
The '2' is in the "hundreds"
The '0' is in the "tens"
The '4' is in the "ones"
.
The '7' is in the "tenths"
the ' 5' is the "hundredths"
So, therefore the value of '2' is in the "hundreds"
Answer:
x = 12
General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
- Equality Properties
Step-by-step explanation:
<u>Step 1: Define equation</u>
-2(x - 4) = -16
<u>Step 2: Solve for </u><em><u>x</u></em>
- Distribute -2: -2x + 8 = -16
- Isolate <em>x</em> term: -2x = -24
- Isolate <em>x</em>: x = 12
<u>Step 3: Check</u>
<em>Plug in x to verify it's a solution.</em>
- Substitute: -2(12 - 4) = -16
- Subtract: -2(8) = -16
- Multiply: -16 = -16
Here we see that -16 does indeed equal -16.
∴ x = 12 is a solution of the equation.
Answer:
31 bags
Step-by-step explanation:
Lets see the total number of pounds he mixed with nuts, peanuts, and chocolate chips.
There is a total of 10 and 1/3 pounds in total. To see how many 1/3rd bag he can have, we have to divide the total by 1/3. So we have:
So, there can be 31 bags
The paraboloid meets the x-y plane when x²+y²=9. A circle of radius 3, centre origin.
<span>Use cylindrical coordinates (r,θ,z) so paraboloid becomes z = 9−r² and f = 5r²z. </span>
<span>If F is the mean of f over the region R then F ∫ (R)dV = ∫ (R)fdV </span>
<span>∫ (R)dV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] rdrdθdz </span>
<span>= ∫∫ [θ=0,2π, r=0,3] r(9−r²)drdθ = ∫ [θ=0,2π] { (9/2)3² − (1/4)3⁴} dθ = 81π/2 </span>
<span>∫ (R)fdV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] 5r²z.rdrdθdz </span>
<span>= 5∫∫ [θ=0,2π, r=0,3] ½r³{ (9−r²)² − 0 } drdθ </span>
<span>= (5/2)∫∫ [θ=0,2π, r=0,3] { 81r³ − 18r⁵ + r⁷} drdθ </span>
<span>= (5/2)∫ [θ=0,2π] { (81/4)3⁴− (3)3⁶+ (1/8)3⁸} dθ = 10935π/8 </span>
<span>∴ F = 10935π/8 ÷ 81π/2 = 135/4</span>
Answer:
1/12
Step-by-step explanation:
1/2 ÷ 6
is the same as;
1/2 × 1/6 = 1/12
