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elena-s [515]
3 years ago
10

Considering the two tables below called Orders and OrderLines, how many rows would be returned by this SQL query? SELECT o.Order

ID, ol.OrderLineID FROM Orders AS o LEFT JOIN OrderLines AS ol ON o.OrderID = ol.OrderID ;

Engineering
1 answer:
natima [27]3 years ago
3 0

Answer:

The query will return four (4) rows

Explanation:

(See attachment below for the table)

The query will return rows where ORDERID is the same on both tables i.e. table ORDERS and ORDERLINES.

The similar rows (on column ORDERID) are 1897 and 87; for table ORDERLINESS and table ORDERS.

But on table ORDERLINESS, there are three rows with 1897 (under ORDERID column).

In total, they have 4 rows in common which are

ORDERID

1897

1897

1897 and

87

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Solve using Matlab the problems:
Firlakuza [10]

Answer:

Explanation:

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clear; clc

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n = input('Total number of objects: ');

r = input('Size of subgroup: ');

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p = 1;

for i = n - r + 1 : n

   p = p*i;

end

str1 = [num2str(p) ' permutations'];

disp(str1)

% Computes and displays combinations according to basic formulas

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How many permutations and combinations can be made of the 15 alphabets, taking four at a time?

The answer is:

32760 permutations

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7 0
2 years ago
Tech A says that horsepower is a measurement simply of the amount of work being performed. Tech B says that horsepower can be ca
snow_tiger [21]

Answer:

Tech B

Explanation:

Horsepower (hp) refers to a unit of measurement of power in respect of the output of engines or motors.

Horsepower is the common unit of power. It indicates the rate at which work is done.

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So,

Tech B is correct

6 0
3 years ago
The coefficient of static friction for both wedge surfaces is μw=0.4 and that between the 27-kg concrete block and the β=20° inc
balandron [24]

Assuming  the wedge has an angle of 5°.The minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.

<h3>Minimum value of force P</h3>

First step

Using this formula to find the weight of the block

W=mg

W=27×9.81

W=264.87 N

Second step

Angles of friction ∅A and ∅B

∅A=tan^-1(μA)

∅A=tan^-1(0.70)

∅A=34.99°

∅B=tan^-1(μB)

∅B=tan^-1(0.40)

∅B=21.80°

Third step

Equate the sum of forces in m-direction to 0 in order to find the reaction force at B.

∑fm=0

W sin (∅A+20°)  + RB cos (∅B+∅A)=0

264.87 sin(34.99°+20°) + RB cos (21.80°+34.99°)=0

216.94+0.5477Rb=0

RB=216.94/0.5477

RB=396.09 N

Fourth step

Equate the sum of forces in x-direction to 0 in order to find force Rc.

∑fx=0

RB cos (∅B) - RC cos (∅B+ 5°)=0

396.09 cos(21.80°) - RC cos (21.80°+5°)=0

RC=396.09 cos(21.80°)/cos(26.80°)

RC=412.02 N

Last step

Equate the sum of forces in y-direction to 0 in order to find force P required to move the block up the incline.

∑fy=0

RB sin (∅B) + RC sin (∅B)-P=0

P=Rb sin (∅B) + RC sin (5°+∅B)

P=396.09 sin(21.80°) +412.02sin (5°+21.80°)

P=322.84 N

Inconclusion the minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.

Learn more about Minimum value of force P here:brainly.com/question/20522149

7 0
2 years ago
Using leftover paint colors is acceptable in a paint shop and will help cut down on waste.
Luden [163]
Im pretty sure it is true !
3 0
3 years ago
Read 2 more answers
1. Calculate the battery life in years when a pacemaker has the following characteristics: Battery Ampere-hours = 1.5 Pulse volt
Wittaler [7]

Answer:

battery life in year = 9 years and 48 days

Explanation:

given data

Battery Ampere-hours = 1.5

Pulse voltage = 2 V

Pulse width = 1.5 m sec

Pulse time period = 1 sec

Electrode heart resistance = 150 Ω

Current drain on the battery = 1.25 µA

to find out

battery life in years

solution

we get first here duty cycle that is express as

duty cycle = \frac{width}{period}      ...............1

duty cycle = 1.5 × 10^{-3}

and applied voltage will be

applied voltage = duty energy × voltage    ...........2

applied voltage = 1.5 × 10^{-3} × 2

applied voltage = 3 mV

so current will be

current = \frac{applied\ voltage}{resistance}   ................3

current = \frac{3}{150}

current = 20 µA

so net current will be

net current = 20 - 1.25

net current = 18.75 µA

so battery life will be

battery life = \frac{1.5}{18.75*10^{-6}}

battery life = 80000 hours

battery life in year = \frac{80000}{8760}

battery life in year = 9.13 years

battery life in year = 9 years and 48 days

4 0
3 years ago
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