Answer
Length measured with Ruler A= 1.17 dm
Length measured with Ruler B= 1.18 dm
Step-by-step explanation:
I included a missing image that shows the pencil's length as measured using two rulers.
Ruler A has a scale marked at 1 dm intervals.
Ruler B has a scale marked at 0.1 dm intervals.
Length measured with Ruler A= 1.17 dm
(The middle markings reveal that the pencil tip lies between 1 and 1.25, slightly above 1.125 but below 1.25, positioning it around 1.2 and 1.125 dm, approximately half of 1.25, which is 1.125, plus an increment of one of the three halves of that half, which gives 1.125+0.045 dm thus concluding to 1.17 dm.)
Length measured with Ruler B= 1.18 dm
(The pencil tip appears just shy of 1.2 dm, confirming the precise value is clearly 1.18 and not 1.19 dm.)
Tip: When unsure about the exact measure, it’s advisable to estimate one extra decimal beyond what the ruler presents.
The digit 5 represents 50 because it is in the tens place.
The customer’s monthly bill approximately amounts to $35.70. By multiplying 35.70 by 0.70 (30% as a decimal), you find 24.99, which is the target amount he aims for now.
The paraboloid intersects the x-y plane when x²+y²=9, defining a circle with a radius of 3, centered at the origin.
<span>Utilizing cylindrical coordinates (r,θ,z), the paraboloid transforms into z = 9−r² and f = 5r²z. </span>
<span>If F represents the average of f over the area 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>Thus, F = 10935π/8 ÷ 81π/2 = 135/4</span>
Answer:
Indeed, the equation is solvable by factoring. By applying the given equation, you can take the square root of both sides. Since both 169 and 9 are perfect squares, the left-hand side simplifies to plus or minus 13/3, producing rational results. Adding 6 to 13/3 yields a rational number while subtracting it does too. Thus, a quadratic equation is factorable if its solutions are rational.
