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Paladinen [302]
3 years ago
14

Please help with this urgently​

Mathematics
2 answers:
tatiyna3 years ago
7 0
It’s picture is blank
.........
Elena-2011 [213]3 years ago
7 0

Answer:

497 by 2 + (x+18) × 1+34 =587

497 by 2 + (x+18) + 34 = 587

497 by 2 + x + 18 + 34 = 587

497 by 2 + x + 52 = 587

601 + 2 x = 1174

2 x = 1174 - 601

2 x = 573

x = 573 by 2

x = 286 1 by 2 , x = 286.5

Step-by-step explanation:

by = /

you can write it as

497

___

2

or

1

_

2

You might be interested in
If there are 3 basketballs for every 7 students at Wilder, then how many students are there if the store room has 49 basketballs
Doss [256]

Answer:

113 is correct because 49 divided by 3 is 16.33. 16.33 multiplied by 7 is 113

Step-by-step explanation:

3 0
3 years ago
3. The following sets are not equal - An (BUC) and AU(BnC) Construct a universe U and non-empty sets A, B, and C so that the abo
yarga [219]

Answer:

1.U={1,2,3,4,5}

A={2}

B={2,3}

C={4,5}

2.U={1,2,3,4}

A={1,2}

B={2,3}

C={4}

Step-by-step explanation:

We are given that A\cap (B\cup C) and A\cup (B\cap C)

are different sets

1.We have to construct a universe set U and non empty sets A,B and C so that above set in fact the same

Suppose U={1,2,3,4,5}

A={2}

B={2,3}

C={4,5}

B\cap C=\phi

B\cup C={2,3,4,5}

A\cap (B\cup C)={2}\cap{2,3,4,5}={2}

A\cup (B\cap C)={2}\cup\phi={2}

Hence, A\cap (B\cup C)=A\cup (B\cap C)

2.We have to construct a universe set U and non empty sets A,B and C so that  above sets are in fact different

Suppose U={1,2,3,4}

A={1,2}

B={2,3}

C={4}

B\cap C=\phi

B\cup C={2,3,4}

A\cup (B\cap C)={1,2}\cup \phi={1,2}

A\cap (B\cup C)={1,2}\cap {2,3,4}={2}

Hence, A\cap (B\cup C)\neq A\cup (B\cap C)

3 0
3 years ago
I got the first two can you help me with the last one PLZ
Nat2105 [25]

1. Geometric Sequence

2. a_n = a_{n-1} * 3

3. a_n = 6 * (3)^{n-1}

Step-by-step explanation:

Given sequence is:

6, 18, 54, 162,....

Here

a_1 =6\\a_2 = 18\\a_3 = 54

(a) Is this an arithmetic or geometric sequence?

We can see that the difference between the terms is not same so it cannot be an arithmetic sequence.

We have to check for common ratio (ratio between consecutive terms of a sequence) denoted by r

r = \frac{a_2}{a_1} = \frac{18}{6}= 3\\r = \frac{a_3}{a_2} = \frac{54}{18} = 3

As the common ratio is same, the given sequence is a geometric sequence.

(b) How can you find the next number in the sequence?

Recursive formulas are used to find the next number in sequence using previous term

Recursive formula for a geometric sequence is given by:

a_n = a_{n-1} * r

In case of given sequence,

a_n = a_{n-1} * 3

So to find the 5th term

a_5 = a_4*3\\a_5 = 162*3\\a_5 = 486

(c) Give the rule you would use to find the 20th week.

In order to find the pushups for 20th week, explicit formul for sequence will be used.

The general form of explicit formula is given by:

a_n = a_1 * r^{n-1}

Putting the values of a_1 and r

a_n = 6 * (3)^{n-1}

Hence,

1. Geometric Sequence

2. a_n = a_{n-1} * 3

3. a_n = 6 * (3)^{n-1}

Keywords: Geometric sequence, common ratio

Learn more about geometric sequence at:

  • brainly.com/question/10666510
  • brainly.com/question/10699220

#LearnwithBrainly

4 0
4 years ago
Find the 4th term (x-y)^12
maria [59]

Answer:

The fourth term of the expansion is -220 * x^9 * y^3

Step-by-step explanation:

Question:

Find the fourth term in (x-y)^12

Solution:

Notation: "n choose k", or combination of k objects from n objects,

C(n,k) = n! / ( k! (n-k)! )

For example, C(12,4) = 12! / (4! 8!) = 495

Using the binomial expansion formula

(a+b)^n

= C(n,0)a^n + C(n,1)a^(n-1)b + C(n,2)a^(n-2)b^2 + C(n,3)a^(n-3)b^3 + C(n,4)a^(n-4)b^4 +....+C(n,n)b^n

For (x-y)^12, n=12, k=3, a=x, b=-y, and the fourth term is

C(n,3)a^(n-3)b^3

=C(12,3) * x^(12-3) * (-y)^(3)

= 220*x^9*(-y)^3

= -220 * x^9 * y^3

3 0
4 years ago
Help<br> thanks!!!!!!!!!!!!!!!!!!!!!!!!!
harkovskaia [24]
The answer is A because that information they gave to you
6 0
3 years ago
Read 2 more answers
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