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jek_recluse [69]
3 years ago
15

What is the slope of the line that contains the points (3, 4) and (1, 2)?

Mathematics
1 answer:
Nostrana [21]3 years ago
7 0
Slope = (y2 - y1) / (x2 - x1)
(3,4)...x1 = 3 and y1 = 4
(1,2)...x2 = 1 and y2 = 2
now we sub
slope = (2 - 4) / (1 - 3)
slope = -2/-2
slope = 1
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Answer: 1) t(n)=0.6(2)^n

2) f(x)=10(5)^x

Step-by-step explanation:

1) Let the function that shows the thickness of the paper after n folds,

t(n) = ab^n         ---------(1)

Since, According to the question,

Initially the thickness of the paper = 0.6

That is, at n = 0, t(0) = 0.6

By equation (1),

0.6 = a(b)^0\implies 0.6 = a

Hence the function that shows the given situation,

t(n) = 0.6 b^n       -----------(2)

Again when we fold the paper the thickness of the paper will be doubled.

Thus, at n = 1, t(1) = 1.2

By equation (2),

1.2 = 0.6 b^1\implies 2 = b

Thus, the complete function is,

t(n) = 0.6 (2)^n    

2) Let the function that is passing through the points (-2, 2/5) and (-1,2),

f(x) = ab^x         ---------(1)

For f(x) = 2, x = -1

By equation (1),

2= ab^{-1}       ---------(2)

Also, For f(x) = 2/5, x = -2

Again, By equation (1),

\frac{2}{5}= a(b)^{-2}

\implies \frac{2}{5}=ab^{-1}b^{-1}=2b^{-1}

\implies \frac{2}{5}=\frac{2}{b}

\implies 2b=10

\implies b = 5

By substituting this value in equation (2),

We get, a = 10

Hence, from equation (1), the function that is passing through the points (-2, 2/5) and (-1,2),

f(x) = 10(5)^x

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