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Maru [420]
3 years ago
8

Find the value of k: 7k÷3=21

Mathematics
2 answers:
lilavasa [31]3 years ago
6 0
Hope this helps. All you want to do is multi or divide to isolate k.

Free_Kalibri [48]3 years ago
3 0
The answer is 9. 7 times 9 is 63 and then 63 divided by 3 is 9.
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Solve the following inequality 5x+12+3x<36
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Answer:

x<3

Step-by-step explanation:

5x+12+3x<36

First subtract 12 from both sides.

5x+3x<24

Next combine like terms on the left side (5x and 3x)

8x<24

Next divide 8 from both sides

x<3

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3 years ago
How many distinct factors does 75 have?
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Step-by-step explanation:

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2 years ago
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When a breeding group of animals is introduced into a restricted area such as a wildlife reserve, the population can be expected
jasenka [17]

Answer:

A. Initially, there were 12 deer.

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. After 15 years, there will be 410 deer.

D. The deer population incresed by 30 specimens.

Step-by-step explanation:

N=\frac{12.36}{0.03+0.55^t}

The amount of deer that were initally in the reserve corresponds to the value of N when t=0

N=\frac{12.36}{0.33+0.55^0}

N=\frac{12.36}{0.03+1} =\frac{12.36}{1.03} = 12

A. Initially, there were 12 deer.

B. N(10)=\frac{12.36}{0.03 + 0.55^t} =\frac{12.36}{0.03 + 0.0025}=\frac{12.36}{y}=380

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. N(15)=\frac{12.36}{0.03+0.55^15}=\frac{12.36}{0.03 + 0.00013}=\frac{12.36}{0.03013}= 410

C. After 15 years, there will be 410 deer.

D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:

ΔN=N(15)-N(10)

ΔN=410 deer - 380 deer

ΔN= 30 deer.

D. The deer population incresed by 30 specimens.

8 0
3 years ago
Pls help. I'm stuck on this question
Sav [38]
547.66666666666666667
7 0
3 years ago
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