All categories

3 years ago

What is the inverse of this function? f(x)=-1/2(x+3)^1/2, x>= -3

Engineering

1 answer:

Black_prince [1.1K]3 years ago

5 0

Answer:

“We need to accept that we won’t always make the right decisions, that we’ll mess up royally sometimes – understanding that failure is not the opposite of success, it’s part of success.” – Arianna Huffington

Explanation:

You might be interested in

Quadrilateral ABCD is a rectangle.<br> If m ZADB = 7k + 60 and mZCDB = -5k + 40, find mZCBD.

yuradex [85]

Hope this helps...........

8 0

3 years ago

Which of these is the BEST description of

strojnjashka [21]

Answer:

i would say C but i may be wrong have a great day

Explanation:

3 0

3 years ago

Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe

NemiM [27]

Answer:

(a) attached below

(b) $pf_{C}=0.85$ $lagging$

(d) $X_{C} =49.37$ Ω

(e) $I_{cap} =9.72 A$ and $I_{line} =27.66 A$

Explanation:

Given data:

$P_{1}=15 kW$

$S_{2} =10 kVA$

$pf_{1} =0.6$ $lagging$

$pf_{2}=0.8$ $leading$

$V=480$ $Volts$

(a) Draw the power triangle for each load and for the combined load.

$\alpha_{1}=cos^{-1} (0.6)=53.13$ °

$\alpha_{2}=cos^{-1} (0.8)=36.86$ °

$S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA$

$Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99$ ≅ $20kVAR$

$P_{2} =S_{2}*pf_{2} =10*0.8=8 kW$

$Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99$ ≅ $-6 kVAR$

The negative sign means that the load 2 is providing reactive power rather than consuming

Then the combined load will be

$P_{c} =P_{1} +P_{2} =15+8=23 kW$

$Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR$

(b) Determine the power factor of the combined load and state whether lagging or leading.

$S_{c} =P_{c} +jQ_{c} =23+14j$

or in the polar form

$S_{c} =26.92$ °

$pf_{C}=cos(31.32) =0.85$ $lagging$

The relationship between Apparent power S and Current I is

$S=VI^{*}$

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

Current of the combined load can be found by

$I_{C} =S_{C}/\sqrt{3}*V$

$I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A$

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

$Q_{C} =3*V^2/X_{C}$

$X_{C} =3*V^2/Q_{C}$

$X_{C} =3*(480)^2/14*10^3$ Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is

$I_{cap} =V/X_{C} =480/49.37=9.72 A$

Line current flowing from the source is

$I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A$

8 0

3 years ago

solniwko [45]

Answer:

Common Uses: Boxwood is well-suited for carving and turning, and the tree's diminutive size restricts it to smaller projects. Some common uses for Boxwood include: carvings, chess pieces, musical instruments (flutes, recorders, woodwinds, etc.), rulers, handles, turned objects, and other small specialty items.If you want a small, compact, low-growing shrub to form a hedge that serves as an accent or border along your walkway, fence line or planting beds, dwarf boxwood varieties are the best pick. The "Dwarf English" boxwood (Buxus sempervirens “Suffruticosa”) creates a border hedge approximately 1 to 2 feet in height.

Explanation:

3 0

3 years ago

Why is tubing sometimes coiled when installed in a car or vechile

andriy [413]

Answer:

Coiled tubing is often used to carry out operations similar to wire lining.

8 0

3 years ago