3/12 = y/8 solve for y

Which table shows a decreasing linear relationship?

Kisachek [45]

Step-by-step explanation:

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3 0

2 years ago

a group of students attend a math club half the students are boys and 4/9 of the boys have brown eyes what fraction of the group

enyata [817]

From the whole group half the club are boys.
fraction of boys from the whole class = 1/2
from this fraction, 4/9th of the boys have brown eyes
therefore fraction of boys who have brown eyes
= 4/9 of 1/2
the 'of' function is used to multiply
$\frac{4}{9} * \frac{1}{2}$
when multiplying fractions numerators are multiplied by numerators and denominators multiplied by denominators
$\frac{4*1}{9*2} = \frac{4}{18}$
4/18 can be simplified by dividing the whole number by a common number
4/18 divided by 2 = 2/9
the answer is 2/9
2/9 of the club are boys with brown eyes

3 0

3 years ago

Read 2 more answers

Using the Breadth-First Search Algorithm, determine the minimum number of edges that it would require to reach

jekas [21]

Answer:

The algorithm is given below.

#include <iostream>

#include <vector>

#include <utility>

#include <algorithm>

using namespace std;

const int MAX = 1e4 + 5;

int id[MAX], nodes, edges;

pair <long long, pair<int, int> > p[MAX];

void initialize()

{

for(int i = 0;i < MAX;++i)

id[i] = i;

}

int root(int x)

{

while(id[x] != x)

{

id[x] = id[id[x]];

x = id[x];

}

return x;

}

void union1(int x, int y)

{

int p = root(x);

int q = root(y);

id[p] = id[q];

}

long long kruskal(pair<long long, pair<int, int> > p[])

{

int x, y;

long long cost, minimumCost = 0;

for(int i = 0;i < edges;++i)

{

// Selecting edges one by one in increasing order from the beginning

x = p[i].second.first;

y = p[i].second.second;

cost = p[i].first;

// Check if the selected edge is creating a cycle or not

if(root(x) != root(y))

{

minimumCost += cost;

union1(x, y);

}

return minimumCost;

}

int main()

{

int x, y;

long long weight, cost, minimumCost;

initialize();

cin >> nodes >> edges;

for(int i = 0;i < edges;++i)

{

cin >> x >> y >> weight;

p[i] = make_pair(weight, make_pair(x, y));

}

// Sort the edges in the ascending order

sort(p, p + edges);

minimumCost = kruskal(p);

cout << minimumCost << endl;

return 0;

}

8 0

2 years ago

What value of x makes the equation true?

Rasek [7]

<em>Greetings from Brasil...</em>

Let's add all the values on one side and make it equal to the sum of all the other values on the other side

(- X) + (- X) + (- X) + (- X) + (- X) + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = X + X + X + (- 1) + (- 1) + (- 1) + (- 1) + (- 1) + (- 1) + (- 1) + (- 1)

- 5X + 12 = 3X - 8

- 5X - 3X = - 8 - 12

- 8X = - 20 x(- 1)

8X = 20

X = 20/8

3 0

2 years ago

Which expression is equivalent to 6X equals 10 X +20

Andrei [34K]

I think it’s 2(5x +10)

7 0

2 years ago