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Dimas [21]
3 years ago
13

Big thank you if you help me ;D

Mathematics
1 answer:
Georgia [21]3 years ago
5 0

Answer:

Step-by-step explanation:

I find it hard to believe that you were asked to find the solution to a cubic by any other method than graphing it, which I have given below. Not only that, the numbers are very awkward.

The best you could do I think is equate the two equations after you have looked at the graph.

The x value must satisfy

0.5(x - 4) = 2x^3 - 2x - 3

0.5x -  2 = 2x^3 - 2x - 3

2x^3 + 0.5x - 2x - 3 +2 = 0

3x^3 - 1.5x - 1 = 0

from which you can get y

My calculator gives the solution as

x =   0.5

y = - 1.75

I have corrected the mistake I made doing this. The answer is now correct. The graph shows the original two equations which agrees with the combined equation given above.

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Step-by-step explanation:

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How do you solve<br> -.8x + 40 - 8x - 35
makvit [3.9K]

Answer:

-16x + 5

Step-by-step explanation:

-8x + 40 -8x - 35

Collect like terms

-8x - 8x + 40 - 35

= - 16x + 5  or 5 - 16x

I hope this was helpful, Please mark as brainliest

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2 3/4 divided by ( -1 2/3
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Step-by-step explanation:

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8 0
2 years ago
How do you find the HCF of 15t and 6s?
earnstyle [38]

Answer:

HCF=3

Step-by-step explanation:

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4 0
3 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
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