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goblinko [34]
3 years ago
14

Assume the population of wolves in a wilderness reserve area increases by the same number from month to month due to migration.

After 12 months, there were 190 wolves; after 25 months, there were 398 wolves. How many wolves will be in the wilderness area after 36 months?
7840

574

590

1400
Mathematics
1 answer:
alekssr [168]3 years ago
8 0
The answer is 574.

We can find the amount of wolves gained every month by dividing the amount by the number of months that have passed.

Net wolf gain per month is 16.


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PLZ HELP!
AnnyKZ [126]

Answer:

$0.25

Step-by-step explanation:

Site A pays ...

... $0.65 × 5 = $3.25

Site B pays ...

... y = 0.60·5 = 3.00 . . . dollars

Site A pays more by ...

... $3.25 - 3.00 = $0.25

_____

You can also look at the difference in unit rates. Site A pays $0.65 per photo, while site B pays $0.60 per photo, a difference of $0.65 - 0.60 = $0.05 per photo. Then 5 photos will be worth $0.05 × 5 = $0.25 more at Site A.

4 0
3 years ago
5x-29>-34 OR 2x+31<29
Margarita [4]
5x-29>-34
5x>-5
x>-1
x<1
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6 0
3 years ago
Read 2 more answers
5. Solve the equation z^2 + 4z - 9 = 0 by completing the square.
Luba_88 [7]

Answer:

A. Z = 2 + N13

Step-by-step explanation:

Let's solve your equation step-by-step.

0=z2+4z−9

Step 1: Subtract z^2+4z-9 from both sides.

0−(z2+4z−9)=z2+4z−9−(z2+4z−9)

−z2−4z+9=0

For this equation: a=-1, b=-4, c=9

−1z2+−4z+9=0

Step 2: Use quadratic formula with a=-1, b=-4, c=9.

z=

−b±√b2−4ac

2a

z=

−(−4)±√(−4)2−4(−1)(9)

2(−1)

z=

4±√52

−2

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8 0
3 years ago
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Zolol [24]
61 is the prime number.

39 is divisible by 3 and 13.
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7 0
3 years ago
Read 2 more answers
Does anybody know the answer to these questions?
san4es73 [151]

Answer:

  1. 625,000 J

  2. 100 J

  4. 5 kg

  5. √5 ≈ 2.236 m/s

Step-by-step explanation:

You should be aware that the SI derived units of Joules are equivalent to kg·m²/s².

To reduce confusion between <em>m</em> for mass and m for meters, we'll use an <em>italic m</em> for mass.

In each case, the "find" variable is what's left after we put the numbers into the formula. It is what the question is asking for. The "given" values are the ones in the problem statement and are the values we put into the formula. The formula is the same in every case.

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1. KE = (1/2)<em>m</em>v² = (1/2)(2000 kg)(25 m/s)² = 625,000 kg·m²/s² = 625,000 J

__

2. KE = (1/2)<em>m</em>v² = (1/2)(0.5 kg)(20 m/s)² = 100 kg·m²/s² = 100 J

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4. KE = (1/2)<em>m</em>v²

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  (250 kg·m²/s²)/(50 m²/s²) = <em>m</em> = 5 kg

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5. KE = (1/2)<em>m</em>v²

  2000 kg·m²/s² = (1/2)(800 kg)v²

  (2000 kg·m²/s²)/(400 kg) = v² = 5 m²/s²

  v = √5 m/s ≈ 2.236 m/s

7 0
3 years ago
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