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trasher [3.6K]
2 years ago
13

Using a Table of Values to Find a Solution to a System

Mathematics
2 answers:
julia-pushkina [17]2 years ago
5 0

Answer:

f(x) = 2.5x - 10.5 and g(x) = 64(0.5)^x

When x = 2,

f(x) = 2.5(2) - 10.5 = -5.5,

g(x) = 64 * 0.5^(2) = 16.

When x = 3,

f(x) = 2.5(3) - 10.5 = -3,

g(x) = 64 * 0.5^(3) = 8.

When x = 4,

f(x) = 2.5(4) - 10.5 = -0.5,

g(x) = 64 * 0.5^(4) = 4.

When x = 5,

f(x) = 2.5(5) - 10.5 = 2,

g(x) = 64 * 0.5^(5) = 2.

When x = 6,

f(x) = 2.5(6) - 10.5 = 4.5,

g(x) = 64 * 0.5^(6) = 1.

Therefore we can see that f(x) = g(x) = 2 when x = 5.

Hence the solution is x = 5.

Step-by-step explanation:

laila [671]2 years ago
4 0

Answer:

This problem wants you to substitute the answers from the table into the equation to find the solution, which is when the input gives the same output basically.

f(x) = 2.5x -10.5                          g(x) = 64(0.5)^x

when x= 2...                                when x= 2...

f(2) = 2.5(2) - 10.5                        g(2) = 64(0.5)^2

f(2) = 5 - 10.5 = -5.5                     g(2) = 64(.25)= 16

when x= 3...                                 when x= 3...

f(3) = 2.5(3) - 10.5                         g(3) = 64(0.5)^3

f(3) = 7.5 - 10.5 = -3                      g(3) = 64(0.125) =8

when x= 4...                               when x = 4...

f(4) = 2.5(4) - 10.5                       g(4) = 64(0.5)^4

f(4) = 10 - 10.5 = -0.5                  g(4) = 64(0.0625) = 4

when x= 5...                               when x= 5...

f(5) = 2.5(5) - 10.5                      g(5) = 64(0.5)^5

f(5) = 12.5 - 10.5 = 2                   g(5) = 64(0.03125) = 2

when x= 6...                               when x = 6...

f(6) = 2.5(6) - 10.5                       g(6) = 64(0.5)^6

f(6) = 15 - 10.5 = 4.5                   g(6) = 64(0.015625) = 1

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Tan(x-y) if cscx=13/5 and coty=4/3
ehidna [41]

Answer:

tan(x-y) = -16/63

Step-by-step explanation

Tan(x-y) if cscx=13/5 and coty=4/3

Given

coty = 4

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1/sinx = 13/5

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opp = 5

hyp = 13

Get the adjacent

hyp² = opp²+adj²

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tan(x-y) = (5-9/12)/1+5/16

tan(x-y) = -4/12/(21/16)

tan(x-y) = -1/3 * 16/21

tan(x-y) = -16/63

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