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1 year ago

A line passes through (6, 3), (8, 4), and (n, -2). Find the value of n.

Mathematics

1 answer:

timurjin [86]1 year ago

8 0

Answer:

n = -4

Step-by-step explanation:

Notice how in the previous examples they all divide by 2? It is the same case here. So now the question is... "what" divided by "2" is "-2".

Let me show you:

n ÷ 2 = -2

if you know your integer math, you should know that...

-4 ÷2 = -2 (this is the correct equation)

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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. N(10) 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. N(10) 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

Evaluate the expression when n= 3. n2+8n+6

Katyanochek1 [597]

Answer:

Step-by-step explanation:

4 0

3 years ago

8th grade khan academy

satela [25.4K]

Answer:

d = 11t - 20

Step-by-step explanation:

Hope it helps and have a great day! =D

~sunshine~

4 0

3 years ago

What is the answer to f-7/g= h for f

gtnhenbr [62]

Answer:

Step-by-step explanation:

$f - \frac{7}{g}=h\\\\f =h + \frac{7}{g}$

5 0

3 years ago

Exhibit 6-2 the weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 poun

Evgen [1.6K]

The question is asking for the lower bound of the 95% two tailed Confidence interval of the normally distributed population.

95% C.I. is given by 200 + or - 1.96(25) = 200 + or - 49 = (151, 249)
Therefore, the minimum weight of the middle 95% of players is 151 pounds.

5 0

3 years ago