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3 years ago

Find a recursive rule for the nth term of the sequence. 5, 20, 80, 320, ...

Mathematics

1 answer:

Ira Lisetskai [31]3 years ago

7 0

Answer:

$a_1 = 5$

$a_n = 4a_{n - 1}$

Step-by-step explanation:

$a_1 = 5$

20/5 = 4

80/20 = 4

320/80 = 4

This a geometric sequence with r = 4.

$a_n = 4a_{n - 1}$

Answer:

$a_1 = 5$

$a_n = 4a_{n - 1}$

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What is the reason for each step in the solution of the equation?<br> -62 +72 +14=3

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3 years ago

What is the circumference of a circle with a radius of 56 inches

DIA [1.3K]

Answer:

351.8584

Step-by-step explanation:

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4 years ago

Find the number of elements in A 1 ∪ A 2 ∪ A 3 if there are 200 elements in A 1 , 1000 in A 2 , and 5, 000 in A 3 if (a) A 1 ⊆ A

lina2011 [118]

Answer:

a. 4600

b. 6200

c. 6193

Step-by-step explanation:

Let $n(A)$ the number of elements in A.

Remember, the number of elements in $A_1 \cup A_2 \cup A_3$ satisfies

$n(A_1 \cup A_2 \cup A_3)=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)$

Then,

a) If $A_1\subseteq A_2, n(A_1 \cap A_2)=n(A_1)=200$ , and if $A_2\subseteq A_3, n(A_2\cap A_3)=n(A_2)=1000$

Since $A_1\subseteq A_2\; and \; A_2\subseteq A_3, \; then \; A_1\cap A_2 \cap A_3= A_1$

$n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-200-200-1000-200=4600$

b) Since the sets are pairwise disjoint

$n(A_1 \cup A_2 \cup A_3)=\\n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\200+1000+5000-0-0-0-0=6200$

c) Since there are two elements in common to each pair of sets and one element in all three sets, then

$n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-2-2-2-1=6193$

8 0

3 years ago

Find the diagonal of a square whose sides are of the given measure.<br><br> Given =

Contact [7]

Answer:

7 $\sqrt{42}$

Step-by-step explanation:

The diagonal is the hypotenuse of the right triangle with legs 7 $\sqrt{3}$

Use Pythagoras' identity to solve for the diagonal d

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is

d² = (7 $\sqrt{3}$ )² + (7 $\sqrt{3}$ )²

= 147 + 147

= 294 ( take the square root of both sides )

d = $\sqrt{294}$ = 7 $\sqrt{42}$ ≈ 45.37 ( to 2 dec. places )

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3 years ago

The sum of three consecutive integers is 915. what are the integers

ra1l [238]

(x-1)+x+(x+1)=915
3x=915
x=305

304, 305, 306

4 0

3 years ago