1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
VashaNatasha [74]
4 years ago
14

 Can you solve 2^x=e^(x+2)

Mathematics
2 answers:
mina [271]4 years ago
8 0
<u>Answer:</u> Yes, I can.


Although you haven't asked for the solution, here it is anyway:

2^x = e^(x+2)

x ln(2) = x+2

x ln(2) - x = 2

x [ ln(2) - 1 ] = 2

x = 2 / [ ln(2) - 1 ]

x = 2 / -0.3069... =<u> - 6.518</u>... (rounded) 

serg [7]4 years ago
7 0
2^x=e^{x+2}\\&#10;\\&#10;ln(2^x)=ln(e^{x+2})\\&#10;\\&#10;xln(2)=(x+2)ln(e)\\&#10;\\&#10;xln(2)=x+2\\&#10;\\&#10;\frac{x+2}{x}=ln(2)\\&#10;\\&#10;\frac{x}{x}+\frac{2}{x}=ln(2)\\&#10;\\&#10;1+\frac{2}{x}=ln(2)\\&#10;\\&#10;\frac{2}{x}=ln(2)-1\\&#10;\\&#10;\boxed{x=\frac{2}{ln(2)-1}}
You might be interested in
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()

* to the API, which returns an int identifier

*/

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) {

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));

}

}

}

8 0
3 years ago
What is the recursive rule for the sequence -22.7, -18.4, -14.1, -9.8, -5.5
nikklg [1K]
The answer would be d.) An=An+1-4.3
6 0
4 years ago
Read 2 more answers
The weight of infants follow a bell shaped distribution with mean
viva [34]
Ummmmmmmmmmmnmmmmmmmmnnnn maybe
4 0
3 years ago
PLEASE HELP FOR 10 POINTS what is the length of PM
Nataly_w [17]

Answer:

D

Step-by-step explanation:

The figures are the same just flipped. We can tell AD=NM ON=AB, BC=OP. Thus DC =MP

DC has the length of 6 and because MP is congruent to DC it also has 6.

8 0
3 years ago
) Determine the probability that a bit string of length 10 contains exactly 4 or 5 ones.
yanalaym [24]

Answer: 0.4512

Step-by-step explanation:

A bit string is sequence of bits (it only contains 0 and 1).

We assume that the  0 and 1 area equally likely to any place.

i.e. P(0)= P(1)= \dfrac{1}{2}

The length of bits : n = 10

Let X = Number of getting ones.

Then , X \sim Bin(n=10,\ p=\dfrac{1}{2})

Binomial distribution formula : P(X=x)=^nC_x p^x q^{n-x} , where p= probability of getting success in each event and q= probability of getting failure in each event.

Here , p=q=\dfrac{1}{2}

Then ,The probability that a bit string of length 10 contains exactly 4 or 5 ones.

P(X= 4\ or\ 5)=P(x=4)+P(x=5)\\\\=^{10}C_4(\dfrac{1}{2})^{10}+^{10}C_4(\dfrac{1}{2})^{10}

=\dfrac{10!}{4!6!}(\dfrac{1}{2})^{10}+\dfrac{10!}{5!5!}(\dfrac{1}{2})^{10}

=(\dfrac{1}{2})^{10}(\dfrac{10!}{4!6!}+\dfrac{10!}{5!5!})

=(\dfrac{1}{2})^{10}(210+252)

=(0.0009765625)(462)

=0.451171875\approx0.4512

Hence, the  probability that a bit string of length 10 contains exactly 4 or 5 ones is 0.4512.

3 0
4 years ago
Other questions:
  • State if the triangle is congruent
    6·1 answer
  • What is the sum of 2.5 x 10^4 and 3.4 c 10^4
    9·2 answers
  • Which equations are related equations to x + 5 = 34?
    12·1 answer
  • Which number is irrational?? Help..!
    9·1 answer
  • A pair of shoes was $5 but is now on sale for<br> $4. What is the percent discount?
    11·2 answers
  • Can someone help me answer this please?
    14·2 answers
  • Por favor alquien me ayuda
    8·1 answer
  • X+y =783<br>10x+7y=6459​
    6·1 answer
  • Q. 16 please help to solve
    14·1 answer
  • Answer The Question (SUPER HARD!!!)<br><br> 2 - 1 + 1 = <br><br> 10 - 8 =<br><br> 1 + 2 - 1 =
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!