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 ;
1 answer:
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
You might be interested in
Answer:
vB = - 0.176 m/s (↓-)
Explanation:
Given
(AB) = 0.75 m
(AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
vB = ?
We use the formulas
Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)
⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m
Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)
⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB
⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)
then we have
⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)
⇒ vB = - 0.176 m/s (↓-)
The pic can show the question.
