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slega [8]
2 years ago
15

I need help I redone this test I failed it the first time and need help right now please​

Mathematics
1 answer:
Eddi Din [679]2 years ago
3 0
3 x 1.5 = 4.5 and positive, negative
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What are the coordinates of A" if it is reflected over the y-axis?
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Answer:

(3,1)

Step-by-step explanation:

When reflecting over the y-axis if the point was on quadrant 1, then the coordinated must be changed like so (x,y) -> (-x,y), so therefore in the scenario, all we have to do is reverse this process. So in this case, Point A is (-3,1) so all we have to do is change the -3 to a positive 3 to the the coordinates of A'' which is (3,1).

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125 cubes

Step-by-step explanation:

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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

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countsearch = countsearch+1;

End if

While (count < n)

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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

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// Print result

cout<<search;

return 0;

}

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3 years ago
What is the answer plzzzzzz help a kid out
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Answer:

the answer is 8 by one of this question.

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