Answer:
C. 905 in³
Step-by-step explanation:
This problem bothers on the mensuration of solid shapes, sphere
We know that the volume of a sphere is expresses as
V= 4/3πr³
Given that the diameter of the sphere is 12inches
Radius r= 12/2= 6 inches
Substituting our radius we can solve for the volume
V=4/3(3.142*6³)
V= (4*3.142*216)/3
V= 2714.68/3
V= 904.9in³
Approximately V= 905in³
<span>Let a_0 = 100, the first payment. Every subsequent payment is the prior payment, times 1.1. In order to represent that, let a_n be the term in question. The term before it is a_n-1. So a_n = 1.1 * a_n-1. This means that a_19 = 1.1*a_18, a_18 = 1.1*a_17, etc. To find the sum of your first 20 payments, this sum is equal to a_0+a_1+a_2+...+a_19. a_1 = 1.1*a_0, so a_2 = 1.1*(1.1*a_0) = (1.1)^2 * a_0, a_3 = 1.1*a_2 = (1.1)^3*a_3, and so on. So the sum can be reduced to S = a_0 * (1+ 1.1 + 1.1^2 + 1.1^3 + ... + 1.1^19) which is approximately $5727.50</span>
<span>Use the formula: r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ] where k = 0,1,2,3,4
</span><span>First 5th root:
k = 0
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*0)/5 ) + i*sin( (280+360*0)/5 ) ]
2*[ cos( (280+360*0)/5 ) + i*sin( (280+360*0)/5 ) ]
2*[ cos( (280+0)/5 ) + i*sin( (280+0)/5 ) ]
2*[ cos( 280/5 ) + i*sin( 280/5 ) ]
2*[ cos( 56 ) + i*sin( 56 ) ]
-------------------------------------------------------------------
Second 5th root:
k = 1
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*1)/5 ) + i*sin( (280+360*1)/5 ) ]
2*[ cos( (280+360*1)/5 ) + i*sin( (280+360*1)/5 ) ]
2*[ cos( (280+360)/5 ) + i*sin( (280+360)/5 ) ]
2*[ cos( 640/5 ) + i*sin( 640/5 ) ]
2*[ cos( 128 ) + i*sin( 128 ) ]
-------------------------------------------------------------------
Third 5th root:
k = 2
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*2)/5 ) + i*sin( (280+360*2)/5 ) ]
2*[ cos( (280+360*2)/5 ) + i*sin( (280+360*2)/5 ) ]
2*[ cos( (280+720)/5 ) + i*sin( (280+720)/5 ) ]
2*[ cos( 1000/5 ) + i*sin( 1000/5 ) ]
2*[ cos( 200 ) + i*sin( 200 ) ]
-------------------------------------------------------------------
Fourth 5th root:
k = 3
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*3)/5 ) + i*sin( (280+360*3)/5 ) ]
2*[ cos( (280+360*3)/5 ) + i*sin( (280+360*3)/5 ) ]
2*[ cos( (280+1080)/5 ) + i*sin( (280+1080)/5 ) ]
2*[ cos( 1360/5 ) + i*sin( 1360/5 ) ]
2*[ cos( 272 ) + i*sin( 272 ) ]
-------------------------------------------------------------------
Fifth 5th root:
k = 4
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*4)/5 ) + i*sin( (280+360*4)/5 ) ]
2*[ cos( (280+360*4)/5 ) + i*sin( (280+360*4)/5 ) ]
2*[ cos( (280+1440)/5 ) + i*sin( (280+1440)/5 ) ]
2*[ cos( 1720/5 ) + i*sin( 1720/5 ) ]
2*[ cos( 344 ) + i*sin( 344 ) ]</span>
Unfortunately, you failed to include in here the amount that has to be paid by the family initially. However, as a general knowledge, the amount that has to be paid for a credit after n years in a simple interest can be calculated through the equation below,
F = P + Prn
where P is the initial value, r is the decimal equivalent of the given rate and n is the number of years. If the given simple interest rate is yearly then, the equation would become,
F = P x (1 + (0.12)(3/12)) = 1.03P
Lets find the missing sides first
big triangle, we apply the Pythagorean theorem a^2+b^2=c^2
a^=13^2-5^=169-25=144, Square root of a=square root of 144=12
a=12
smaller triangle
5^2-3^2=b^2
25-16=9
b=4
Elm trail is 13km+5km=18km
Dogwood trail=12+5+4+3=24km
correct choices are 2 and 5
