Answer:
6283 in³
Step-by-step explanation:
The largest sphere that can fit into the cardboard box must have its diameter, d equal to the length, L of the cardboard box.
Since the cardboard box is in the shape of a cube, its volume V = L³
So, L = ∛V
Since V = 12000 in³,
L = ∛(12000 in³)
L= 22.89 in
So, the volume of the sphere, V' = 4πr³/3 where r = radius of cube = L/2
So, V = 4π(L/2)³/3
= 4πL³/8 × 3
= πL³/2 × 3
= πL³/6
= πV/6
= π12000/6
= 2000π
= 6283.19 in³
≅ 6283.2 in³
= 6283 in³ to the nearest whole cubic inch
Answer: 212 7/20
Step-by-step explanation: We know that 212 is the whole number. So we need to convert .35 into a fraction in simplest form. Any decimal that has a hundredths place will be divided by 100 to get the fraction. In this case that fraction would be 35/100. And to convert that into simplest form, you need to find the least (or lowest) common denominator which is 5. Then you divide the numerator and denominator by 5 and get 7/20, so your answer is 212 7/20
Answer:
The degree is 6, and the zero is 0
Step-by-step explanation:
Hope this helps! :) ~Zane
P.S. sorry if im wrong with the zero one
{{{ THE BOLDED CHARACTERS SHOULD BE SMALL. }}}
SEQUENCE: 6, 18, 54, 162
18/6 = 3
54/18 = 3
162/54 = 3
then, r (common ratio) = 3
_________________________________________
RECURSIVE RULE: r = 3
an = a(n - 1) × r [formula]
ANSWER: an = a(n - 1) × 3
_________________________________________
ITERIATIVE RULE: r = 3, a1 = 6
an = a1 × r^(n - 1) [formula] [ ^(n-1) is an exponent]
ANSWER: an = 6 × 3^(n - 1)
Answer:
perimeters of the rectangle=p=46.014 metres
Step-by-step explanation:
Given that:
Length (l) = 21 m
Area of rectangle(A) = 42.15 meter-square
Width (w)=?
Required data:
Perimater of Rectangle=p=?
Calculation:
As we know that Area of rectangle=A=l*w
Putting the value we get
42.15 m(square)=(21 m)*w
or w=42.15/21
or w=2.007 m
Now to find perimters of rectangle we know that
p=2(l + w) metres
putting the values
p=2(21+2.007) metres
p=2(23.007) metres
p=46.014 metres