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Margaret [11]
3 years ago
13

What is the initial value in y=14000.(0.92)​

Mathematics
1 answer:
Anettt [7]3 years ago
7 0

Answer:

12,880

Step-by-step explanation:

this is exponential decay

a=i(1-r)

it should be 14000

But if you forgot to write a variable on the right side of the equation that that would change things.

The decay rate however would be 8%, since 1-.92 is .8

the new equation would be a=14000(1-0.8)

you can also continue solving for y

y=12,880

when you just multiply

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Answer:

Option C

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Step-by-step explanation:

Given: 500x^3 +108y^\left (18\right )

the common factor from 500x^3 and 108y^\left (18\right ) is 4.

therefore, 4\cdot \left ( 125x^3+27y^\left (18\right ) \right )

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Now, use the formula for above expression: (a^3+b^3)=(a+b)(a^2-ab+b^2)

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( (5x)^3+(3y^6)^3 \right))=(5x+3y^6)(25x^2-15xy^6+9y^12)

Therefore, we have

500x^3 +108y^\left (18\right )=4\cdot (5x+3y^6)\left (25x^2-15xy^6+9y^\left ( 12 \right )  \right )

Therefore, one of the factors of 500x^3 +108y^\left (18\right ) is (5x+3y^6).








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