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dexar [7]
3 years ago
5

A biologist studied the populations of white-sided jackrabbits and black-tailed jackrabbits over a 5-year

Mathematics
1 answer:
natita [175]3 years ago
8 0
The correct answer is B
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Factor −1/4 out of −1/2x−5/4y. The factored expression is .
Olegator [25]

Answer:

  (-1/4)(2x +5y)

Step-by-step explanation:

Multiply and divide by -1/4:

  -\frac{1}{2}x-\frac{5}{4}y=\dfrac{(-\frac{1}{4})(-\frac{1}{2}x-\frac{5}{4}y)}{-\frac{1}{4}}=\left(-\dfrac{1}{4}\right)(-4)\left(-\dfrac{1}{2}x-\dfrac{5}{4}y\right)\\\\=\left(-\dfrac{1}{4}\right)\left(2x+5y)

7 0
3 years ago
You are deciding between two cell phone plans. Plan A: Unlimited text, talk, and data for $50 per month. Plan B: Unlimited text
Sergio039 [100]
Sent a picture of the solution to the problem (s).

7 0
3 years ago
Jenna has a rectangular pig pen that measures 8m wide by 10m long. She wants to increase the area to 168 m^2 by increasing the l
babunello [35]

Answer:

The answer is A

Step-by-step explanation:

I just took the test and it was the answer!

4 0
3 years ago
a total eclipse in 2003 lasted 2 17/24,minutes in 2005 it lasted 3/8 of a minute how much longer did it last in 2003
Doss [256]

Answer:

Total eclipse was 2\frac{1}{3} minutes longer than in 2005.

Step-by-step explanation:

It is given in the question that total eclipse in 2003 lasted 2\frac{17}{24} minutes and in 2005 it lasted 3/8 of a minute.

We have to calculate how much longer did it last in 2003.

So we have to subtract the duration in 2003 by duration in 2005

Total time taken in 2003- total time taken in 2005

=2\frac{17}{24}-\frac{3}{8}

\frac{48+17}{24}-\frac{3}{8}

=\frac{65}{24}-\frac{3}{8}

=\frac{65-9}{24}

=\frac{56}{24}

\frac{7}{3}

=2\frac{1}{3} minutes


3 0
3 years ago
Read 2 more answers
Annie buys a picnic table priced at $407. If the sales tax is 6%, what will be the total cost including tax?
Oksi-84 [34.3K]

Answer:

-$407

-tax:6%

407 times 6 divided by 100

413/100

4.13

407 plus 4.13

$411.13

3 0
2 years ago
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