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Scorpion4ik [409]
2 years ago
4

Write a SELECT statement that returns these columns from the Invoices table: The invoice_number column The invoice_date column T

he invoice_date column plus 30 days The payment_date column A column named days_to_pay that shows the number of days between the invoice date and the payment date The number of the invoice date’s month The four-digit year of the invoice date When you have this working, add a WHERE clause that retrieves just the invoices for the month of May based on the invoice date, not the number of the invoice month
Engineering
1 answer:
Eduardwww [97]2 years ago
7 0

This query would work on an Oracle database:

SELECT

invoice_number,

invoice_date,

invoice_date+30,

payment_date,

payment_date-invoice_date AS days_to_pay,

to_date(invoice_date, 'MM'),

to_date(invoice_date, 'YYYY')

FROM Invoices

WHERE to_char(to_date(invoice_date, 'MM')) = '05'

In fact, the operation with dates are allowed there (i.e. date+1 means "the day after the date"), and date1-date2 return the days between the two dates.

You might be interested in
(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5
bezimeni [28]

Answer:

a.) 1.453MW/m2,  b.)  2,477,933.33 BTU/hr  c.) 22,733.33 BTU/hr  d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

                        50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, Q = \frac{109*0.5*100}{0.0075} = 726,666.667W

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, Q=\frac{1*0.5*100}{0.0075} = 6,666.67W

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, Q=\frac{109*0.5*100}{0.015} =363,333.33W

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0
2 years ago
Tech A says that as moisture levels increase in brake fluid, the boiling point of the brake fluid decreases. Tech B says that fi
dezoksy [38]

Answer: Both Technician A and B are correct.

Explanation:

Technicians A and B are both right about their diagnosis. The Society of Automotive Engineers performed extensive research on vehicle brake fluids and found that there is typically a 2% moisture content in the brake fluid after a year of operating a vehicle. And as the moisture content of the brake fluid rises, the boiling point of the brake fluid decreases as well.

4 0
2 years ago
A 150-lbm astronaut took his bathroom scale (a spring scale) and a beam scale (compares masses) to the moon where the local grav
kozerog [31]

Answer:

a)Wt =25.68 lbf

b)Wt = 150 lbf

F= 899.59 N

Explanation:

Given that

g = 5.48 ft/s^2.

m= 150 lbm

a)

Weight on the spring scale(Wt) = m g

We know that

1\ lbf=32.17 \ lmb.ft/s^2

Wt = 150 x 5.48/32 lbf

Wt =25.68 lbf

b)

On the beam scale

This is scale which does not affects by gravitational acceleration.So the wight on the beam scale will be 150 lbf.

Wt = 150 lbf

If the plane is moving upward with acceleration 6 g's then the for F

F = m a

We know that

1\ ft/s^2= 0.304\ m/s^2

5.48\ ft/s^2= 1.66\ m/s^2

a=6 g's

a=9.99\ m/s^2

So

F = 90 x 9.99 N

F= 899.59 N

3 0
2 years ago
A steel rotating-beam test specimen has an ultimate strength Sut of 1600 MPa. Estimate the life (N) of the specimen if it is tes
ziro4ka [17]

Answer:

the life (N) of the specimen is 46400 cycles

Explanation:

given data

ultimate strength Su = 1600 MPa

stress amplitude σa = 900 MPa

to find out

life (N) of the specimen

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su   .............1

Se = 0.5 × 1600

Se = 800 Mpa

and we know

Se for steel is 700 Mpa for Su ≥ 1400 Mpa

so we take endurance limit Se is = 700 Mpa

and strength of friction f  = 0.77 for 232 ksi

because for Se 0.5 Su at 10^{6} cycle = (1600 × 0.145 ksi ) = 232

so here coefficient value (a) will be

a = \frac{(f*Su)^2}{Se}    

a = \frac{(0.77*1600)^2}{700}  

a = 2168.3 Mpa

so

coefficient value (b) will be

a = -\frac{1}{3}log\frac{(f*Su)}{Se}

b =  -\frac{1}{3}log\frac{(0.77*1600)}{700}

b = -0.0818

so no of cycle N is

N =  (\frac{ \sigma a}{a})^{1/b}

put here value

N =  (\frac{ 900}{2168.3})^{1/-0.0818}

N = 46400

the life (N) of the specimen is 46400 cycles

5 0
2 years ago
To water his lawn, a homeowner uses two hoses. One connects to the faucet, the other to the end of the first hose to make the ho
Shtirlitz [24]

Answer: to be exact you need 28mm of tubing for that

Explanation:

When the election

8 0
2 years ago
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