Answer: Increase in the population of lower organisms.
Explanation: Raccoons are mesopredators found in the middle levels of food webs and have critical impacts on the dynamics of many other species. Thus, they greatly contribute to the functioning of the ecosystem.
They are naturally equipped to checkmate the population size of other organism that are below them (i.e., their preys) in the food web. They are; predators, pathogen carriers (such as rabies), and competitors, as they compete with some specialist in the food web.
Hence, the decrease in the population of raccoons will lead to the increase in the population size of lower organisms they prey on in the food web or chain.
A crane is set up for steel erection at the site of a 5 story office building where each story is 15 feet tall. The new building will be 50 feet wide and 75 feet long and the crane is located 5 feet off the southwesterly corner of the building. Assuming that you need a 10 foot boom clearance over the last piece of steel.
Required:
Compute the minimum boom length
Answer:
127.58 ft
Explanation:
We need first to calculate the length from corner to corner of the story, L.
Since the length of each floor is 75 ft and its width 50 ft, and since each floor is a rectangle with diagonal, L, using Pythagoras' theorem, we have
L² = (75 ft)² + (50 ft)²
= 5625 ft² + 2500 ft²
L² = 8125 ft²
L = √(8125 ft²)
L = 90.14 ft
Since the crane is 5 ft off the southwesterly corner of the building, the working radius, R = L + 5 ft = 90.14 ft + 5 ft = 95.14 ft.(since the diagonal length of the floor plus the distance of the crane from the south westerly corner add to give the working radius)
The boom tip height, H = height of building h + clearance of boom, h'
h = height of each story, h" × number of stories, n
Since h" = 15 ft and n = 5
h = 15 ft × 5 = 75 ft
Also, h' = 10 ft
So, H = h + h'
H = 75 ft + 10 ft
H = 85 ft
So, the minimum boom length, L' = √(H² + R²)
substituting the values of the variables into the equation, we have
L' = √(H² + R²)
L' = √((85 ft)² + (95.14 ft)²)
L' = √(7225 ft² + 9051.6196 ft²)
L' = √(16276.6196 ft²)
L' = 127.58 ft
"All data types are keywords but all keywords are not data types", justify this statement.
Keywords Are Reserved Words Which Are Not Use As Normal Cases But Used For Special Purpose In Our Program whereas, Data Type Tells The Compiler And Interpreter How The Program Is Going To Be Executed And Can Be Used At Every Cases.
Hope This Helps You ❤️Based on how they are stored in the computer, the types produced by these keywords can be categorized into two families, integer types and floating-point types.
What is the relation between data and keyword?Bit, byte, and word are terminology that can be used to refer to either memory or data storage units in computers.
The main distinction between text fields and keyword fields is that text fields are examined during indexing, whereas keyword fields are not.
This means that while keyword fields are indexed exactly as is, text fields are split down into their constituent phrases at indexing to allow for partial matching.
A word or collection of words that an Internet user enters into a search engine or search bar to conduct a search is referred to as a keyword in the context of digital marketing.
Therefore, All data types are keywords, but all keywords are not data types.
Learn more about data and keywords here:
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SN1 reactions usually proceed with ________. A) equal amounts of inversion and retention at the center undergoing substitution B) slightly more inversion than retention at the center undergoing substitution C) slightly more retention then inversion at the center undergoing substitution D) complete inversion at the center undergoing substitution E) complete retention at the center undergoing substitution complete retention at the center undergoing substitution slightly more retention then inversion at the center undergoing substitution complete inversion at the center undergoing substitution equal amounts of inversion and retention at the center undergoing substitution slightly more inversion than retention at the center undergoing substitution
Answer:
equal amounts of inversion and retention at the center undergoing substitution
Explanation:
In an SN1 reaction, the rate determining step is the formation of a carbonation which is flat and planar.
This means that both faces of the carbo cation are equally available for attack by the nucleophile.
Attack on either of the faces may occur equally thereby yielding a racemic mixture.
in c the square root of a number N can be approximated by repeated calculation using the formula NG = 0.5(LG + N/LG) where NG stands for next guess and LG stands for last guess. Write a function that calculates the square root of a number using this method. The initial guess will be the starting value of LG. The program will com- pute a value for NG using the formula given. The difference between NG and LG is checked to see whether these two guesses are almost identical. If they are, NG is accepted as the square root; otherwise, the next guess (NG) becomes the last guess (LG) and the process is repeated (another value is computed for NG, the difference is checked, and so on). The loop should be repeated until the difference is less than 0. 005. Use an initial guess of 1. 0. Write a driver function and test your square root function for the numbers 4, 120. 5, 88, 36.01, 10,000, and 0. 25
PLEASE İN C PROGRAMMİNG
Answer:
Following are the program to the given question:
#include <stdio.h>//header file
double square_root(double N, double initialGuess)//defining a method square_root that takes two variable in parameters
{
double NG, LG = initialGuess,diff;//defining double variable
while(1)//use loop to calculate square root value
{
NG = 0.5 * (LG + N / LG);//using given formula
diff = NG - LG;//calculating difference
if(diff < 0)//use if to check difference is less than 0
diff = -diff;//decreaing difference
if(diff < 0.005)//use if that check difference is less than 0.005
break;//using break keyword
else//defining else block
{
LG = NG;//holding value
}
}
return NG;//return value
}
int main()//defining main method
{
double ans, n,initialguess = 1.0;//defining double variable
n = 4;//use n to hold value
ans = square_root(n, initialguess);//calculating the square root value and print its value
printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number
n = 120.5;//use n to hold value
ans = square_root(n, initialguess);//calculating the square root value and print its value
printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number
n = 36.01;//use n to hold value
ans = square_root(n, initialguess);//calculating the square root value and print its value
printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number
n = 0.25;//use n to hold value
ans = square_root(n, initialguess);//calculating the square root value and print its value
printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number
printf("\nEnter a number: ");//print message
scanf("%lf", &n);//input value
ans = square_root(n, initialguess);//calculating the square root value and print its value
printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number
}
Output:
Please find the attachment file.
Explanation:
In this code, a method "square_root" is declared that takes two variable "N, initialGuess" in its parameters, inside the method a three double variable is declared.It uses the given formula and uses the diff variable to hold its value and uses two if to check its value is less than 0 and 0.005 and return its calculated value.In the main method, three double variables are declared that use the "n" to hold value and "ans" to call the method that holds its value and print its value.Question 4. Select the statement that best describes the PM wave. a. Carrier phase angle changes represent the PM modulated signal, however, unlike FM and AM modulated carriers, it is not always easy to see these phase changes in the plot. b. The FM signal accurately represents the message. c. Although PM and FM techniques are classified as angular modulation techniques, they cannot be used interchangeably (i.e., modulated with FM and demodulated with PM, and visa versa). d. All statements are correct.
Answer:
c. Although PM and FM techniques are classified as angular modulation techniques, they cannot be used interchangeably i.e., modulated with FM and demodulated with PM, and vice versa.
Explanation:
Phase modulation is modulation pattern for conditioning communication signals for transmission. PM signals appears to change frequency with message wave. Both PM and FM are angular modulation techniques. The message in PM wave is captured in phase changes.
