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sertanlavr [38]
3 years ago
8

Find the height of a paralleogram, if its area is 84square inches, its base is 12 inches, and its height is respresented by x +

2.

Mathematics
2 answers:
Ghella [55]3 years ago
4 0
Hope you are having clear concept on this.

pentagon [3]3 years ago
3 0

\boxed {\text {Area of Parallelogram  = base } \times \text{height}}

<u>Given area = 84 in², base = 12 in and height = (x + 2) in:</u>

12 (x + 2) = 84

-

<u>Find x:</u>

12 (x + 2) = 84

12x + 24 = 84

12x = 84 - 24

12x = 60

x = 5 inches

-

<u>Find height:</u>

Height = x + 2 = 5 + 2 = 7 inches

-

Answer: Height = 7 inches

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Using linked lists or a resizing array; develop a weighted quick-union implementation that removes the restriction on needing th
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Answer:

Step-by-step explanation:

package net.qiguang.algorithms.C1_Fundamentals.S5_CaseStudyUnionFind;

import java.util.Random;

/**

* 1.5.20 Dynamic growth.

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* removes the restriction on needing the number of objects ahead of time. Add a method newSite()

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*/

public class Exercise_1_5_20 {

public static class WeightedQuickUnionUF {

private int[] parent; // parent[i] = parent of i

private int[] size; // size[i] = number of sites in subtree rooted at i

private int count; // number of components

int N; // number of items

public WeightedQuickUnionUF() {

N = 0;

count = 0;

parent = new int[4];

size = new int[4];

}

private void resize(int n) {

int[] parentCopy = new int[n];

int[] sizeCopy = new int[n];

for (int i = 0; i < count; i++) {

parentCopy[i] = parent[i];

sizeCopy[i] = size[i];

}

parent = parentCopy;

size = sizeCopy;

}

public int newSite() {

N++;

if (N == parent.length) resize(N * 2);

parent[N - 1] = N - 1;

size[N - 1] = 1;

return N - 1;

}

public int count() {

return count;

}

public int find(int p) {

// Now with path compression

validate(p);

int root = p;

while (root != parent[root]) {

root = parent[root];

}

while (p != root) {

int next = parent[p];

parent[p] = root;

p = next;

}

return p;

}

// validate that p is a valid index

private void validate(int p) {

if (p < 0 || p >= N) {

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}

public boolean connected(int p, int q) {

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}

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int rootP = find(p);

int rootQ = find(q);

if (rootP == rootQ) {

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// make smaller root point to larger one

if (size[rootP] < size[rootQ]) {

parent[rootP] = rootQ;

size[rootQ] += size[rootP];

} else {

parent[rootQ] = rootP;

size[rootP] += size[rootQ];

}

count--;

}

}

public static void main(String[] args) {

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Random r = new Random();

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8 0
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Answer:

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Step-by-step explanation:

This is the answer

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