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viva [34]
2 years ago
5

Each level of a smartphone app adds more gems for you to match. On level one, there were 13 gems. On level twelve, there were 12

3 gems. When you complete all twelve levels, how many total gems will you have matched? A. 605 B. 660 C. 748 D. 816
Computers and Technology
1 answer:
Sidana [21]2 years ago
5 0
<span>Level : 1,2,3,4,5,6,7,8,9,10,11,12
Gems : 13 to 123
We have to increase from 13 to 123 over a span of 11 levels.
That's an increase of 110 over 11 levels.
110 / 11 = 10
So we go up 10 gems each level Level : 1,2,3,4,5,6,7,8,9,10,11,12
Gems : 13 23 33 43 53 63 73 83 93 103 113 123</span>
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a

Explanation:

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Define a structure type auto_t to represent an automobile. Include components for the make and model (strings), the odometer rea
yulyashka [42]

Answer:

see explaination

Explanation:

#include <stdio.h>

#include <string.h>

#define BUFSIZE 1000

struct auto_t scan_auto(char *);

struct date_t {

char day[2];

char month[2];

char year[4];

};

struct tank_t {

char tankCapacity[10];

char currentFuelLevel[10];

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struct auto_t {

char make[50];

char model[50];

char odometerReading[10];

struct date_t manufactureDate;

struct date_t purchaseDate;

struct tank_t gasTank;

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int main(int argc, char *argv[]) {

/* the first command-line parameter is in argv[1]

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/* read in one line, up to BUFSIZE-1 in length */

while(fgets(buff, BUFSIZE - 1, fp) != NULL)

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fclose(fp); /* close the file */

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struct auto_t scan_auto(char *line) {

int spacesCount = 0;

int i, endOfMake, endOfModel, endOfOdometer;

for (i = 0; i < sizeof(line); i++) {

if (line[i] == ' ') {

spacesCount++;

if (spacesCount == 1) {

endOfMake = i;

}

else if (spacesCount == 2) {

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else if (spacesCount == 3) {

endOfOdometer = i;

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struct auto_t newAuto;

int count = 0;

for (i = 0; i < endOfMake; i++) {

newAuto.make[count++] = line[i];

}

newAuto.make[count] = '\0';

count = 0;

for (i = endOfMake+1; i < endOfModel; i++) {

newAuto.model[count++] = line[i];

}

newAuto.model[count] = '\0';

count = 0;

for (i = endOfModel+1; i < endOfOdometer; i++) {

newAuto.odometerReading[count++] = line[i];

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newAuto.odometerReading[count] = '\0';

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8 0
2 years ago
The groups_per_user function receives a dictionary, which contains group names with the list of users. Users can belong to multi
Akimi4 [234]

The groups_per_user function receives a dictionary, which contains group names with the list of users.

Explanation:

The blanks to return a dictionary with the users as keys and a list of their groups as values is shown below :

def groups_per_user(group_dictionary):

   user_groups = {}

   # Go through group_dictionary

   for group,users in group_dictionary.items():

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   return(user_groups)

print(groups_per_user({"local": ["admin", "userA"],

       "public":  ["admin", "userB"],

       "administrator": ["admin"] }))

3 0
2 years ago
Thomas has signed a deal with a production house that allows them to use his images on their website. What is required when imag
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D. Fair use. Because you need to give credit to the owner.
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2 years ago
Read 2 more answers
Network flow issues come up in dealing with natural disasters and other crises, since major unexpected events often require the
nasty-shy [4]

Answer:

Check the explanation

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This approach of sending patients to hospitals doesn't break the capacity limitation of edges. Hence we can send patient "p" to hospital "h" with 1 flow- unit if edge(m uh) permits at least 1 flow- unit.

The running-time of this algorithm is found by finding the time needed to solve max-flow graph with nodes O(n+k) and edges O(n^{k}) edges.

5 0
2 years ago
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