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storchak [24]
3 years ago
11

Whitney has a sprinkler than covers a circular area with diameter 24 feet. What is the area of lawn that the sprinkler covers? (

Use 3.14 for π)
Mathematics
1 answer:
grin007 [14]3 years ago
6 0
Answer:
I think the answer is 7.64331210191

Explanation:
I divided 24 by 3.14 and got 7.6433121091
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Let C(n, k) = the number of k-membered subsets of an n-membered set. Find (a) C(6, k) for k = 0,1,2,...,6 (b) C(7, k) for k = 0,
vladimir1956 [14]

Answer:

(a) C(6,0) = 1, C(6,1) = 6, C(6,2) = 15, C(6,3) = 20, C(6,4) = 15, C(6,5) = 6, C(6,6) = 1.

(b) C(7,0) = 1, C(7,1) = 7, C(7,2) = 21, C(7,3) = 35, C(7,4) = 35, C(7,5) = 21, C(7,6) = 7, C(7,7)=1.

Step-by-step explanation:

In this exercise we only need to recall the formula for C(n,k):

C(n,k) = \frac{n!}{k!(n-k)!}

where the symbol n! is the factorial and means

n! = 1\cdot 2\cdot 3\cdot 4\cdtos (n-1)\cdot n.

By convention 0!=1. The most important property of the factorial is n!=(n-1)!\cdot n, for example 3!=1*2*3=6.

(a) The explanations to the solutions is just the calculations.

  • C(6,0) = \frac{6!}{0!(6-0)!} = \frac{6!}{6!} = 1
  • C(6,1) = \frac{6!}{1!(6-1)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6
  • C(6,2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2\cdot 4!} = \frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15
  • C(6,3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!\cdot 3!} = \frac{5!\cdot 6}{6\cdot 6} = \frac{5!}{6} = \frac{120}{6} = 20
  • C(6,4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!\cdot 2!} = frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15
  • C(6,5) = \frac{6!}{5!(6-5)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6
  • C(6,6) = \frac{6!}{6!(6-6)!} = \frac{6!}{6!} = 1.

(b) The explanations to the solutions is just the calculations.

  • C(7,0) = \frac{7!}{0!(7-0)!} = \frac{7!}{7!} = 1
  • C(7,1) = \frac{7!}{1!(7-1)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7
  • C(7,2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2\cdot 5!} = \frac{6!\cdot 7}{2\cdot 5!} = \frac{5!\cdot 6\cdot 7}{2\cdot 5!} = \frac{6\cdot 7}{2} = 21
  • C(7,3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!\cdot 4!} = \frac{6!\cdot 7}{6\cdot 4!} = \frac{5!\cdot 6\cdot 7}{6\cdot 4!} = \frac{120\cdot 7}{24} = 35
  • C(7,4) = \frac{7!}{4!(7-4)!} = \frac{6!\cdot 7}{4!\cdot 3!} = frac{5!\cdot 6\cdot 7}{4!\cdot 6} = \frac{120\cdot 7}{24} = 35
  • C(7,5) = \frac{7!}{5!(7-2)!} = \frac{7!}{5!\cdot 2!} = 21
  • C(7,6) = \frac{7!}{6!(7-6)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7
  • C(7,7) = \frac{7!}{7!(7-7)!} = \frac{7!}{7!} = 1

For all the calculations just recall that 4! =24 and 5!=120.

6 0
2 years ago
Find the value of the variable and DF if D is between C and F if CD 4y -9, DF 2y-7, and CF 14.
Yuri [45]
CF = CD + DF (since D is simply a point on the line, and C to that point added onto F to that point is CF), so 4y-9+2y-7=14. Adding it up, we get 6y-16=14 and adding by 16 we get 30=6y. Dividing both sides by 6, we get y=5. Plugging that into DF, we get 10-7=DF=3
4 0
3 years ago
In this equation, what is n?
raketka [301]
N is the variable, you solve by subtracting 6 from 20, getting n = 14.
5 0
3 years ago
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A quadratic equation function f(x)=ax^2 passes through (0,0) and (5,5). write an equation for this function
Alex
Equation: f(x)=(1/5)x^2 or f(x)=(0.2)x^2

Check:
f(0)=(1/5)(0)^2
f(0)=(1/5)(0)
f(0)=0✓
f(5)=(1/5)(5)^2
f(5)=(1/5)(25)
f(5)=5✓
6 0
3 years ago
PLEASE HELP IM STUCK
kkurt [141]

Answer:

y= 2x + 10

Step-by-step explanation:

The variable depends on the $2 so, the green box should be 2

4 0
1 year ago
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