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sp2606 [1]
3 years ago
5

Can someone help me with 10 math questions (:

Mathematics
1 answer:
olga2289 [7]3 years ago
3 0
So volue of a cone=1/3 times (area of base[which is a circle]) times height

area of base=area of circle=pi time radius^2
area=pi times 5^2
area=pi times 25=25pi

1/3 times 25pi times 18
25pi times 18 times 1/3
25pi times 18/3
25pi times 6=150pi


the answer is 150π in^3 or 150π cubic inches or C

( to solve, aprox pi to 3.141592 an multiply 150 by 3.141592=471.239 in^3)
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A soup can has a 3 and one eighth in diameter and is 5 in tall. What is the area of the paper that will be used to make the labe
Mamont248 [21]

Answer: 49.09 cm2

Step-by-step explanation:

Hi, to answer this question we have to apply the next formulas:

Circumference of a circle = π x diameter = 3.14159x 3 1/8 = 9.81746 in

The circumference of the can is equal to the length of the rectangular label:

Surface area (S)= circumference(length) x height  

Replacing with the values given:

S = 9.81746 x 5 = 49.09 cm2

Feel free to ask for more if needed or if you did not understand something.

5 0
3 years ago
Plant A is 18 inches tall after one week 36 inches tall after two weeks 56 inches tall after three weeks Plan B is 18 inches tal
KATRIN_1 [288]
So what is your question
8 0
3 years ago
In what situation would you use an ANOVA statistical test over T-test?
expeople1 [14]

Step-by-step explanation:

when means of more than two groups are to be compared, ANOVA is perferred

5 0
2 years ago
You want to put a 5 inch thick layer of topsoil for a new 23 ft by 18 ft garden. the dirt store sells by the cubic yards. how ma
lianna [129]

5 inches = 5/12 = 0.4166 feet

23 x 18 x 0.4166 = 172.5 cubic feet

1 cubic foot = 0.037037 cubic yards

172.5 x 0.037037 = 6.39 round up to 6.5 cubic yards

5 0
3 years ago
Read 2 more answers
Given a term in the geometric sequence and the common ratio find the first five terms the explicit form and the recursive formul
Pani-rosa [81]
A1 = 4
a2 = 5a1 = 5 x 4 = 20
a3 = 5a2 = 5 x 20 = 100
a4 = 5a3 = 5 x 100 = 500
a5 = 5a4 = 5 x 500 = 2,500

Tn = ar^(n-1); where a = 4, r = 5
Tn = 4(5)^(n-1) = 4/5 (5)^n
Explicit formular is Tn = 4/5 (5)^n

Recursive formular is a_{n+1} = 5a_{n-1}
5 0
3 years ago
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