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Luda [366]
3 years ago
12

First real answer i’ll give Brainlyist

Engineering
1 answer:
pochemuha3 years ago
6 0

Answer:

I think this answer is number B

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In the pressure filled driving environment, handicapping yourself by drinking and driving is not a good choice.
vodka [1.7K]

Answer:

True

Explanation:

Please mark me brainlist

6 0
3 years ago
Design roller chain drive. Specify the chain size, the sizes and number of teeth in the sprockets, the number of chain pitches,
Rom4ik [11]

Answer:

The total design can be summarized as follows:

AC motor 55 hp, 750 rpm  as a driver

Service factor 1.0

Design power 5 hp

No. 40 chain, 0.50 in pitch, 1 strands

17 teeth, 2.72 in pitch dia, 1 strand small sprocket

39 teeth, 7.46 in pitch dia, 1 strand large sprocket

1 strand of length 54 in

Center distance of 19.9 in

Actual Output speed of 326.9 rpm

Type B lubrication

Explanation:

4 0
3 years ago
What process does a professional use to find an object’s position with respect to the camera?
shtirl [24]

Answer:

"D

Explanation:

5 0
3 years ago
Read 2 more answers
Using a queue.
beks73 [17]

Answer:

// Radix Sort

#include<iostream>

using namespace std;

// function to get max value

int getMaximum(int array[], int n)

{

int max = array[0];

for (int i = 1; i < n; i++)

if (array[i] > max)

max = array[i];

return max;

}

// function to get min value

int getMinimum(int array[], int n)

{

int mn = array[0];

for (int i = 1; i < n; i++)

if (array[i] <mn)

mn= array[i];

return mn;

}

//counting sort

void counterSort(int array[], int n, int expo)

{

int out[100]; // out max

int i, count[10] = {0};

// Store count of occurrences in count[]

for (i = 0; i < n; i++)

count[ (array[i]/expo)%10 ]++;

// Change count[i] so that count[i] now contains actual

// position of this digit in out[]

for (i = 1; i < 10; i++)

count[i] += count[i - 1];

// construct the out max

for (i = n - 1; i >= 0; i--)

{

out[count[ (array[i]/expo)%10 ] - 1] = array[i];

count[ (array[i]/expo)%10 ]--;

}

for (i = 0; i < n; i++)

array[i] = out[i];

}

void counterSortDesc(int array[], int n, int expo)

{

int out[100]; // out max

int i, count[10] = {0};

// Store count of occurrences in count[]

for (i = 0; i < n; i++)

count[ (array[i]/expo)%10 ]++;

for (i = 10; i >=1; i--)

count[i] += count[i - 1];

// construct out max

for (i = 0; i >= n-1; i++)

{

out[count[ (array[i]/expo)%10 ] - 1] = array[i];

count[ (array[i]/expo)%10 ]++;

}

// Copy the out max to array[], so that array[] now

// contains sorted numbers according to current digit

for (i = 0; i < n; i++)

array[i] = out[i];

}

// Radix Sort function

void radixsort(int array[], int n)

{

// get maximum number

int m = getMaximum(array, n);

for (int expo = 1; m/expo > 0; expo *= 10)

counterSort(array, n, expo);

}

void radixsortDesc(int array[], int n)

{

// get minimum number

int m = getMinimum(array, n);

for (int expo = 1; m/expo > 0; expo *= 10)

counterSortDesc(array, n, expo);

}

// print an max

void print(int array[], int n)

{ cout<<"\n";

for (int i = 0; i < n; i++)

cout << array[i] << " ";

}

// Main function

int main()

{

int array[] = {185, 25, 35, 90, 904, 34, 2, 66};

int n = sizeof(array)/sizeof(array[0]);

radixsort(array , n);

print(array, n);

radixsortDesc(array,n);

print(array,n);

return 0;

}

Explanation:

7 0
3 years ago
XYZ Company purchased a new machine on January 1, 2020 for $160,000. The machine was assigned a 40-year life and a $3,000 residu
AlekseyPX

Answer:

Accumulated depreciation at December 32, 2023 = 29679

Explanation:

machine = 160000

useful life = 40 years

residual value = 3000

straight-line depreciation rate = (160000 - 3000) ÷ 40

                                    = 3925 per year

Annual depreciation rate = 100% ÷ 40 = 2.5%

double-declining rate = 2 × 2.5% = 5%

Depreciation for 2020 = 5% × 160000 = 8000

Depreciation for 2021 = 5% × (160000 - 8000) = 7600

Depreciation for 2022 = 5% × (160000 - 8000 - 7600) = 7220

Depreciation for 2023 = 5% × (160000 - 8000 - 7600 - 7220) = 6859

Accumulated depreciation at December 32, 2023

                    = 8000 + 7600 +7220 + 6859

                    =  29679

4 0
3 years ago
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