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Kay [80]
3 years ago
15

Let f(x) = 4+3z and g(x)=x^2-3. Find each function Value: (f-g)(2)

Mathematics
1 answer:
arlik [135]3 years ago
8 0

Answer:

12z

Step-by-step explanation:

f=(x) = 4+3z and g(x)= x^2-3.

f= (x)7z

g(x) = x^ 2-3

g(x) =1

(f-g) (2)

7z- 1 (2)

6z ×2

12 z

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30°<br> 8 + 6x<br> 4x + 2 <br> in a triangle
andrey2020 [161]

Answer:

x=12

Step-by-step explanation:

Simplifying

30 + 4x + 2 = 8 + 6x

Reorder the terms:

30 + 2 + 4x = 8 + 6x

Combine like terms: 30 + 2 = 32

32 + 4x = 8 + 6x

Solving

32 + 4x = 8 + 6x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-6x' to each side of the equation.

32 + 4x + -6x = 8 + 6x + -6x

Combine like terms: 4x + -6x = -2x

32 + -2x = 8 + 6x + -6x

Combine like terms: 6x + -6x = 0

32 + -2x = 8 + 0

32 + -2x = 8

Add '-32' to each side of the equation.

32 + -32 + -2x = 8 + -32

Combine like terms: 32 + -32 = 0

0 + -2x = 8 + -32

-2x = 8 + -32

Combine like terms: 8 + -32 = -24

-2x = -24

Divide each side by '-2'.

x = 12

Simplifying

x = 12

4 0
3 years ago
a car purchased for $34,000 is expected to lose value, or depreciate, at a rate of 6% per year.using x for years and y for the v
Vika [28.1K]

Answer:

y = 34000(1-0.06) ^ t

After 7.40 years it will be worth less than 21500

Step-by-step explanation:

This problem is solved using a compound interest function.

This function has the following formula:

y = P(1-n) ^ t

Where:

P is the initial price = $ 34,000

n is the depreciation rate = 0.06

t is the elapsed time

The equation that models this situation is:

y = 34000(1-0.06) ^ t

Now we want to know after how many years the car is worth less than $ 21500.

Then we do y = $ 21,500. and we clear t.

21500 = 34000(1-0.06) ^ t\\\\log(21500/34000) = tlog(1-0.06)\\\\t = \frac{log(21500/34000)}{log(1-0.06)}\\\\t = 7.40\ years.

After 7.40 years it will be worth less than 21500

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