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svp [43]
3 years ago
10

Plz don't cheap out on me for points I really need help with this

SAT
1 answer:
Sever21 [200]3 years ago
3 0

The answer fam is......... D)

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Consider the competitive market for sports jackets. The following graph shows the marginal cost (MC), average total cost (ATC),
Orlov [11]

The marginal cost is the additional cost incurred for a extra unit of production.

<h3>How to illustrate the information?</h3>

The average total cost is simply the total cost divided by the number of units.

Based on the table attached, the number of jackets is used to depict whether the producer should shut down or produce.

The short run supply curve is also illustrated.

Learn more about marginal cost on:

brainly.com/question/17230008

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7 0
2 years ago
A humpback whale is classified as a mammal while a basking shark is a fish.
kvasek [131]

The taxonomic classification that is the same for a humpback whale and a basking shark is Phylum.

<h3>What is Taxonomic Classification?</h3>

We want to find Which taxonomy classification are the same for a humpback whale and a basking shark.

A chordate in the animal kingdom is referred to as Kingdom Animalia of the phylum Chordata. This phylum is one that includes all the vertebrates and subphyla tunicates and cephalochordates.

Chordates have peculiar characteristics such as notochord, pharyngeal slits, nerve cord, and a tail.

Therefore, we can conclude that the taxonomic classification that is the same for a humpback whale and a basking shark is Phylum.

Read more about Taxonomic Classification at; brainly.com/question/21023329

8 0
2 years ago
How to add maths??? teacher gave question me to!:
Law Incorporation [45]

Answer:

The answer is 4

Explanation:

7 0
3 years ago
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A 40-kilogram mass is moving across a horizontal surface at 5.0 m/s. what is the magnitude of the net force required to bring th
aliina [53]

Answer:

25 N

Explanation:

Use the equation F = m(Vf–Vi) / Δt.

average force = mass (final velocity - initial velocity) / change in time

F = 40 kg(5.0 m/s) / 8.0 s = 200 / 8.0 = 25 kgm / s^2 = 25 Newtons

4 0
2 years ago
Help me with the question
lorasvet [3.4K]
10 images per day. Since it can receive 3 mb per second for 11 hours a day, that’s up to 118,800 megabits it can receive in one day. By multiplying the amount of gigabits in a typical picture (11.2) by the amount of megabits in a gigabit (1024) you get that there’s 11,468.8 megabits in each picture. Lastly, divide the number of megs that the station receives in one day by the amount of megs in a picture, and you get 10 and some change, therefore it can receive up to ten FULL pictures in a day
8 0
4 years ago
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