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TEA [102]
3 years ago
9

Cd's at sound house are marked 1/3 off. what percent is this

Mathematics
1 answer:
Goshia [24]3 years ago
5 0

1/3=.33333333, which is 33%

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I think of a number, double it and add 14. The result is 36. What is the number?
never [62]

Answer:

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Step-by-step explanation:

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Please help me with my homework
geniusboy [140]

Answer:

A. 5

B. 6

I hope this is helpful for you! Unless my calculator is wrong!

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The diagram show a 5cm x 6cm x 7cm cuboid pls help you
kotegsom [21]

Answer:

d ≈ 10.5

Step-by-step explanation:

The formula for the diagonal of a retangular prism (cuboid) is

d=\sqrt{l^2+w^2+h^2}

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d=\sqrt{5^2+6^2+7^2}

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5 0
3 years ago
Which is the best example of a short-term savings goal?
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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;

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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