Let First Sphere be the Original Sphere
its Radius be : r
We know that Surface Area of the Sphere is : 4π × (radius)²
⇒ Surface Area of the Original Sphere = 4πr²
Given : The Radius of Original Sphere is Doubled
Let the Sphere whose Radius is Doubled be New Sphere
⇒ Surface of the New Sphere = 4π × (2r)² = 4π × 4 × r²
But we know that : 4πr² is the Surface Area of Original Sphere
⇒ Surface of the New Sphere = 4 × Original Sphere
⇒ If the Radius the Sphere is Doubled, the Surface Area would be enlarged by factor : 4
9/5+3/1
24/5 24/5 24/5 24/5 24/5 24/5
Hope this helps! :D
Answer:
D. 96.9 square centimeters
Step-by-step explanation:
A=bh
11.4 x 8.5
=96.9 square centimeters.
Answer: 427 miles
Step-by-step explanation:
if the jeep averages 30.5 miles per gallon, and you have 14 gallons, then you will need to multiply 30.5*14 to get your answer, which is 427
Answer:
![[p-|p|*10^{-3} \, , \, p+|p|* 10^-3]](https://tex.z-dn.net/?f=%5Bp-%7Cp%7C%2A10%5E%7B-3%7D%20%5C%2C%20%2C%20%5C%2C%20p%2B%7Cp%7C%2A%2010%5E-3%5D)
Step-by-step explanation
The relative error is the absolute error divided by the absolute value of p. for an approximation p*, the relative error is
r = |p*-p|/|p|
we want r to be at most 10⁻³, thus
|p*-p|/|p| ≤ 10⁻³
|p*-p| ≤ |p|* 10⁻³
therefore, p*-p should lie in the interval [ - |p| * 10⁻³ , |p| * 10⁻³ ], and as a consecuence, p* should be in the interval [p - |p| * 10⁻³ , p + |p| * 10⁻³ ]
