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d1i1m1o1n [39]
3 years ago
10

One of the factors of the polynomial x3 + 5x2 + 6x is (x2 + 3x). What is the other factor?

Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
7 0

The given polynomial is :

x^{3}+5x^{2}+6x

One factor is given as : x^{2} +3x or x(x+3)

So, now solving this for 2nd factor, we get,            

x(x^{2}+5x+6 )          

x[(x^{2}+3x+2x+6)]          

x(x+3)(x+2)              

Hence, the second factor is (x+2)

labwork [276]3 years ago
6 0
The answer is 
<span>
x (x² + 5x + 6) </span>
<span>-------------------- </span>
<span>x ( x + 3) </span>

<span>( x + 3 ) (x + 2) </span>
<span>--------------------- </span>
<span>x + 3 </span>

<span>x + 2</span>
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8 0
2 years ago
Two trials are conducted for an experiment. In Trial A, the measured value is 240, while the actual value is 200. In Trial B, th
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A is the answer 456% and 512%

Step-by-step explanation:

Percent Errors

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195

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​?

​?

Complete the table.

Percent Errors

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5 0
2 years ago
g Which of the following is not appropriate to use the t-confidence interval for . a. The sample is based on a simple random sam
KATRIN_1 [288]

Answer:

d. The population standard deviation is known.

Step-by-step explanation:

T-distribution or Z-distribution, which to use?

If we have the population standard deviation is known, we use the z-distribution.

Otherwise, if we only have the standard deviation for the sample, we use the t-distribution.

Which of the following is not appropriate to use the t-confidence interval for

When the population standard deviation is known, so letter d.

7 0
2 years ago
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;

if(num >= numbers [I])

{

numbers [I] = num;

}

else

{

goto lbl;

}

}

// Search for a particular number

int search;

cin>>searchdigit;

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

{

if(numbers[I] == searchdigit

search++

}

}

// Print result

cout<<search;

return 0;

}

8 0
3 years ago
Perform the operation and reduce the answer fully. 3/4 + 1/10
Assoli18 [71]

Answer:

17/20

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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