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Oksi-84 [34.3K]
3 years ago
11

If a client is shown the "Cadillac"

Engineering
1 answer:
vivado [14]3 years ago
7 0

Answer:

B

Explanation:

The price is not factored in

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Pls follow me in brainly​
Sergio [31]

Answer:

sure

Add on:

!!

5 0
3 years ago
Read 2 more answers
Use the predicate specifications(x, y): x beats yF (x): x is an (American) football teamQ(x, y): x is quarterback of yL(x,y): x
attashe74 [19]

Answer:

a) ∀y∃x(Q(x, y))

b) (B(Jayhawks, W ildcats)→¬∀y(L(Jayhawks, y)))

c) ∃x(B(Wildcats, x) ∧ B(x, Jayhawks))

Explanation:

a) The statement can be rewritten as "For all football teams, there exists a quarterback" which is written in logical symbols.

b) The statement is an implication and thus have a premise and a conclusion. The premise states "Jayhawks beat the Wildcats" which is translated using B(x, y). The conclusion can be rewritten as "It is not the case that Jayhawks lose to all football teams".

c) The statement is a simple conjunction which can be written as "There exists a team x such that the Wildcats beats x and x beats Jayhawks"

7 0
3 years ago
. . . . . . . . . . ..........
Naddika [18.5K]

Answer: Thx for points...

Explanation:

Heres Random Info About Enginnering (because why not): Engineers are involved in all aspects of interactive TV technology, from designing new cables, to creating new film emulsions, to engineering better sound quality. This technology allows viewers to select any program, film, or game from more than 500 channels. Engineers play an instrumental role in the theme park industry?

Hope This Helps Someday!

5 0
3 years ago
A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust end
hjlf

Answer:

a) v=19.6 m/s

b) H=19.58 m

c) v_{f}=29.57 m/s  

Explanation:

a) Let's calculate the work done by the rocket until the thrust ends.

W=F_{tot}h=(F_{thrust}-mg)h=(35-(2*9.81))*25=384.5 J

But we know the work is equal to change of kinetic energy, so:

W=\Delta K=\frac{1}{2}mv^{2}

v=\sqrt{\frac{2W}{m}}=19.6 m/s

b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.

v_{f}^{2}=v_{i}^{2}-2gh

At the maximum height the velocity is 0, so v(f) = 0.

0=v_{i}^{2}-2gH

H=\frac{19.6^{2}}{2*9.81}=19.58 m  

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.

Using the same equation of part b)

v_{f}^{2}=v_{i}^{2}-2gh

v_{f}=\sqrt{19.6^{2}-(2*9.81*(-25))}=29.57 m/s

The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.

I hope it helps you!

6 0
3 years ago
A ladder logic program and the associated physical input/output components are given below. Lighting changes from full darkness
Katena32 [7]

Answer:

O2 is true.

Explanation:

8 0
4 years ago
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