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alexandr1967 [171]
3 years ago
10

A = (1, 3, 5, 7, 9)

Mathematics
1 answer:
iragen [17]3 years ago
5 0

Answer:

AU(B n C)= (1,3,5,7,9,6)

Step-by-step explanation:

BnC= (6)

AU(Bn C)= (1,3,5,7,9)U(6)

A U(B n C)= (1,3,5,7,9,6)

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Figure ABCD is a trapezoid. Find
Rudiy27

Answer:

x = 3

Step-by-step explanation:

The midsegment is half the sum of the parallel bases, that is

\frac{1}{2} (3x + 3 + 4x + 2) = 13

\frac{1}{2} (7x + 5) = 13 ( multiply both sides by 2 )

7x + 5 = 26 ( subtract 5 from both sides )

7x = 21 ( divide both sides by 7 )

x = 3

3 0
3 years ago
I will literally pay you pls help me :(
vovikov84 [41]

The answer is 35

Since they are supplementary they both equal 180 degrees.

(3x+10)+(2x-5)=180

Add them up

5x-5=180

Subtract 5 from both sides

5x=180-5

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3 years ago
A dime has a radius of about 0.85 cm what is the closest circumfrence
Anuta_ua [19.1K]
5.34 cm. Hope I helped!
4 0
3 years ago
Please help me i need this asap
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Step-by-step explanation:

B is the answer for you. Have a good day!

3 0
3 years ago
The sequence$$1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2,1,2,\dots$$consists of $1$'s separated by blocks of $2$'s with $n$ $2$'s i
kicyunya [14]

Consider the lengths of consecutive 1-2 blocks.

block 1 - 1, 2 - length 2

block 2 - 1, 2, 2 - length 3

block 3 - 1, 2, 2, 2 - length 4

block 4 - 1, 2, 2, 2, 2 - length 5

and so on.


Recall the formula for the sum of consecutive positive integers,

\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2

Now,

1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016

which means that the 1234th term in the sequence occurs somewhere about 1/5 of the way through the 49th 1-2 block.

In the first 48 blocks, the sequence contains 48 copies of 1 and 1 + 2 + 3 + ... + 47 copies of 2, hence they make up a total of

\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224

numbers, and their sum is

\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400

This leaves us with the contribution of the first 10 terms in the 49th block, which consist of one 1 and nine 2s with a sum of 1+9\times2=19.

So, the sum of the first 1234 terms in the sequence is 2419.

8 0
2 years ago
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