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LuckyWell [14K]
3 years ago
12

What type of triangle has no congruent sides and one angle that has a measure greater than 90° but less than 180°?

Mathematics
2 answers:
ololo11 [35]3 years ago
4 0

Answer:

D as it is the obtuse

Step-by-step explanation:

UkoKoshka [18]3 years ago
3 0

Answer:

D.) obtuse scalene triangle

Step-by-step explanation:

acute triangle: 3 acute angles

right triangle: 1 right angle, 2 acute angles

obtuse triangle: 1 obtuse angle, 2 acute angles

scalene triangle: no congruent sides

isosceles triangle: 2 congruent sides

equilateral triangle: 3 congruent sides

The question is:

"What type of triangle has <em>no congruent sides</em> and <u>one angle that has a measure greater than 90° but less than 180°</u>?"

That means <em>scalene</em> and <u>obtuse</u>.

Answer: D.) obtuse scalene triangle

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A highway is going to be built around a city. The highway will be in the shape of a perfect circle, and its total length will be
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Answer:

The area that is surrounded by the highway on the billboard is approximately 89.34 square meters

Step-by-step explanation:

The given parameters are;

The length of the circle shaped highway = 251.2 km

The scale of the map of the bus routes along the highway = 1:7,500

The circumference of a circle, C = 2·π·r = The length of the perimeter of the circle

Therefore, we have;

The circumference of the highway = The length of the circle shaped highway = 251.2 km

The circumference of the highway, C = 251.2 km = 2·π·r

The radius of the circular highway path, r = 251.2 km/(2·π) ≈ 39.994217 km

The area of a circle, A = π·r².

The area of the highway = π·r² = π × (39.994217 km)² ≈ 5,025.09492434 km.²

Given that the scale of the map = 1:7,500. we have;

The scale of the area = (1:7,500)² = (1:56,250,000)

The size of the drawing, 5,025.09492434 km.²/(56,250,000) = 0.00008933502 km² = 89.33502 m.²

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2 years ago
Using linked lists or a resizing array; develop a weighted quick-union implementation that removes the restriction on needing th
balu736 [363]

Answer:

Step-by-step explanation:

package net.qiguang.algorithms.C1_Fundamentals.S5_CaseStudyUnionFind;

import java.util.Random;

/**

* 1.5.20 Dynamic growth.

* Using linked lists or a resizing array, develop a weighted quick-union implementation that

* 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;

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private void validate(int p) {

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

throw new IndexOutOfBoundsException("index " + p + " is not between 0 and " + (N - 1));

}

}

public boolean connected(int p, int q) {

return find(p) == find(q);

}

public void union(int p, int q) {

int rootP = find(p);

int rootQ = find(q);

if (rootP == rootQ) {

return;

}

// 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) {

WeightedQuickUnionUF uf = new WeightedQuickUnionUF();

Random r = new Random();

for (int i = 0; i < 20; i++) {

System.out.printf("\n%2d", uf.newSite());

int p = r.nextInt(i+1);

int q = r.nextInt(i+1);

if (uf.connected(p, q)) continue;

uf.union(p, q);

System.out.printf("%5d-%d", p, q);

uf.union(r.nextInt(i+1), r.nextInt(i+1));

}

}

}

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