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meriva
2 years ago
13

Which of the following is an example of how an abiotic component affects a community.

SAT
1 answer:
Diano4ka-milaya [45]2 years ago
7 0

Answer:

<em><u>A wet spring leads to an increase in the frog population. ..</u></em>

Explanation:

<em>hope</em><em> it</em><em> will</em><em> help</em><em> you</em><em> have</em><em> a</em><em> great</em><em> day</em><em> bye</em>

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Categorize each statement as true or false. The SI V-Scope is a virtual microscope program which simulates the use of a compound
qaws [65]

The SI V-Scope is indeed a virtual microscope program that can stimulate the use of a compound light microscope so this is <u>TRUE</u>.

<h3>What is a SI V-Scope?</h3>

The SI V-Scope stands for a Science Interactive and then Virtual scope which is used simulate the way that a compound light microscope works.

These microscopes are very useful because they help people who don't have a physical microscope to access microscopes wherever they are.

Find out more on virtual microscopes at brainly.com/question/14469908

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3 0
2 years ago
Day and night kennel charges $20 per day plus a one-time food fee of $15 to board a pet. Bark time hotel charges $30 per day plu
Sonbull [250]

The system of equation that represents this situation is y = 20x + 15 for day and night kernel and y = 30x + 5 for bark time hotel.

A linear equation is in the form:

y = mx + b

where y, x are variables, m is the rate of change and b is the initial value of y.

Let y represent the total charges for for boarding a pet in x days.

Day and night kennel charges $20 per day plus a one-time food fee of $15 to board a pet. Hence:

y = 20x + 15

Bark time hotel charges $30 per day plus a one-time food fee of $5. Hence:

y = 30x + 5

The system of equation that represents this situation is y = 20x + 15 for day and night kernel and y = 30x + 5 for bark time hotel.

Find out more at: brainly.com/question/21835898

7 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
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