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Dmitry [639]
2 years ago
7

Hi so I need this answer quickly, and I’m rewarding 15 points

Mathematics
2 answers:
amid [387]2 years ago
7 0

Answer:

It would be 3 1/4

Step-by-step explanation:

your welcome

Alchen [17]2 years ago
4 0
-3.25 is the correct answer
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Why are only some quadrilaterals parallelograms?
borishaifa [10]

A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel . ... Opposite sides are congruent; Adjacent angles are supplementary; The diagonals bisect each other.

3 0
3 years ago
Ryan had 220 baseball cards
emmasim [6.3K]

Answer:

220 baseball cards

Step-by-step explanation:

You didn't tell us anything s o i guess 220 baseball cards

8 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
A certain bacterium splits itself into 2 identical cells in 1 day. Each of those new cells is capable of splitting itself into 2
stiks02 [169]

Answer:

2 with the second power??

Step-by-step explanation:

if you have two with the second power II exponents will multiply it by itself so the number gradually gets larger each time

6 0
3 years ago
Always, sometimes, or never.
irinina [24]
The ratio is always 1:1:√2
3 0
3 years ago
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