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3 years ago

What is the next number of the pattern. 2 7 26 101 400

2 answers:

7 0

$the\ sequence\ can\ be\ determined\ recursively:\\\\ \left \{ {\big{a_1=2\ \ \ \ \ \ \ \ \ \ \ \ \ \ } \atop\big {a_{n+1}=4\cdot a_n-n}} \right. \\\\\\a_6=4\cdot a_5-5=4\cdot400-4=1600-5=1595\\\\Ans.\ the\ nxt\ term\ is\ 1595.$

Paladinen [302]3 years ago

6 0

2x4-1
7x4-2
26x4-3
101x4-4
400x4-5
and so on . .
just do the math.

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Step-by-step explanation please<br> (Worth 3 marks)

atroni [7]

Answer:

62°

Step-by-step explanation:

180-118=62

62+62=124

180-124=56

118-56=62

5 0

3 years ago

Read 2 more answers

Helen had Php7500 for shopping money. When she got home, she had Php132.75 in her pocket. How much did she spend for shopping?

Nataly_w [17]

Answer:

Step-by-step explanation:

Helen had Php7500 for shopping money

she had Php132.75 in her pocket

she spend= 7500.00-132.75

=7367.25answer

4 0

3 years ago

HURRY!

yuradex [85]

-3 x + 1 > y
1/2x -1 (less than or equal to) y

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3 0

2 years ago

An introductory psychology class with n=44 students has 9 freshman males, 15 freshman females, 8 sophomore males, and 12 sophomo

andrezito [222]

Answer:

0.61

Step-by-step explanation:

Pr (female) = total number of females(n')/Total number of students(n)

Where P(female) = probability of selectinga female

Pr(female) = n'/n................. Equation 1

Given: n = 44 students, n' = 15+12 = 27 females

Substitute into equation 1

Pr(female) = 27/44

Pr(female) = 0.61.

Hence the probability of selecting a female is 0.61

3 0

3 years ago

Boris choos three diffrent numbers.the sum of the three numbers is 36.One of trh numbers is a cube number.the other two number a

inna [77]

Answer:

4,5,27

Problem:

Boris chose three different numbers.

The sum of the three numbers is 36.

One of the numbers is a perfect cube.

The other two numbers are factors of 20.

Step-by-step explanation:

Let's pretend those numbers are:

$a,b, \text{ and } c$ .

We are given the sum is 36: $a+b+c=36$ .

One of our numbers is a perfect cube. $a=n^3$ where $n$ is an integer.

The other two numbers are factors of 20. $bk=20$ and $ci=20$ where $a,c,i, \text{ and } k \text{ are integers}$ .

$n^3+\frac{20}{k}+\frac{20}{i}=36$

From here I would just try to find numbers that satisfy the conditions using trial and error.

$3^3+\frac{20}{2}+\frac{20}{2}$

$27+10+10$

$47$

$3^3+\frac{20}{4}+\frac{20}{5}$

$27+5+4$

$36$

So I have found a triple that works:

$27,5,4$

The numbers in ascending order is:

$4,5,27$

4 0

3 years ago