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Rina8888 [55]
3 years ago
8

Factor 64a^2 - 48a + 9. A(8a + 9)(8a - 1) B(8a - 3)^2 C(4a + 3)(16a - 3)

Mathematics
2 answers:
Akimi4 [234]3 years ago
7 0

Answer:

B. (8a-3)^2

Step-by-step explanation:

Hope this helps! :)))

Jobisdone [24]3 years ago
5 0
(8a-3)(8a-3)..........
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The normal probability distribution, with mean M and standard deviation S, can be represented in the following notation.

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1. Let L be a list of numbers in non-decreasing order, and x be a given number. Describe an algorithm that counts the number of
e-lub [12.9K]

Answer:

Algorithm

Start

Int n // To represent the number of array

Input n

Int countsearch = 0

float search

Float [] numbers // To represent an array of non decreasing number

// Input array elements but first Initialise a counter element

Int count = 0, digit

Do

// Check if element to be inserted is the first element

If(count == 0) Then

Input numbers[count]

Else

lbl: Input digit

If(digit > numbers[count-1]) then

numbers[count] = digit

Else

Output "Number must be greater than the previous number"

Goto lbl

Endif

Endif

count = count + 1

While(count<n)

count = 0

// Input element to count

input search

// Begin searching and counting

Do

if(numbers [count] == search)

countsearch = countsearch+1;

End if

While (count < n)

Output count

Program to illustrate the above

// Written in C++

// Comments are used for explanatory purpose

#include<iostream>

using namespace std;

int main()

{

// Variable declaration

float [] numbers;

int n, count;

float num, searchdigit;

cout<<"Number of array elements: ";

cin>> n;

// Enter array element

for(int I = 0; I<n;I++)

{

if(I == 0)

{

cin>>numbers [0]

}

else

{

lbl: cin>>num;

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{

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}

// Search for a particular number

int search;

cin>>searchdigit;

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{

if(numbers[I] == searchdigit

search++

}

}

// Print result

cout<<search;

return 0;

}

8 0
3 years ago
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