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3 years ago

How do you solve this?

Mathematics

1 answer:

Arturiano [62]3 years ago

3 0

Answer:

x=-5

Step-by-step explanation:

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Can someone explain and answer please??

Nat2105 [25]

Simplify the ration expression
$\frac{3x-6}{x-2}$

start by seeing if anything in the numerator or denominator can be factored,
you should notice that you can factor out a 3 from 3x-6 in the numerator

when you do so, the result is
$\frac{3 *(x-2)}{x-2}$

now, since there is an x-2 in the numerator and denominator, they "cancel" out since

$\frac{x-2}{x-2} =1$
however, only when $x \neq 2$
since $\frac{0}{0} \neq 1$

when you cancel out the x-2, you're left with 3

hence, your answer is C

8 0

3 years ago

1) Determine the rate of change for the table below: 2) Determine the equation of the line (using slope-intercept form) that int

agasfer [191]

Answer:

b.adot5

Step-by-step explanation:

7 0

3 years ago

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

3 years ago

A boat took 5 h to travel 60 km up a river, against the current. The return trip took 3 h. Find the speed of the boat in still w

Dafna11 [192]

Answer:

Correct answer: V₁ boat = 16 km/h and V₂ current = 4 km/h

Step-by-step explanation:

If we take speed (velocity) of the boat as V₁ and speed of the current V₂

Let us form equations:

Traveling down the current V₁ + V₂ = 60/3 = 20 km/h

Traveling against the current V₁ - V₂ = 60/5 = 12 km/h

V₁ - V₂ = 12

V₁ + V₂ = 20

If we add up the left and right sides of the equations we get:

2 V₁ = 32 => V₁ = 32/2 = 16 km/h

from the second equation we get:

V₂ = 20 - V₁ = 20 - 16 = 4 km/h

V₂ = 4 km/h

God is with you!!!

8 0

3 years ago

1/5, 2/3, 5/8 from least to greatest on a number line

Svetllana [295]

1/5, then 5/8, then 2/3

8 0

3 years ago