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sveticcg [70]
2 years ago
15

Describe different situations in the real world that could be modeled and solved by a system of equations in two variables.

Mathematics
1 answer:
andrey2020 [161]2 years ago
4 0

solved by a system of equations in two variables.

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Which statement is true about the end behavior of the graphed function?
tatyana61 [14]

The <em><u>correct answer</u></em> is:

A) as the x-values go to positive infinity, the functions values go to negative infinity.  

Explanation:

We can see in the graph that the right hand portion continues downward to negative infinity.  The right hand side of the graph is "as x approaches positive infinity," since x continues to grow larger and larger.  This means as x approaches positive infinity, the value of the function approaches negative infinity.

8 0
3 years ago
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The function f(t)= 15000(1.08)^t models the value of an investment t years from now. What is the meaning of the value of f(10)
guajiro [1.7K]

in the f(t) = 15000(1.08)ᵗ, which is a form of a compounded interest formula, t = years, so


f(t) = 15000(1.08)¹⁰ , is the value of it when t = 10, after 10 years.


4 0
4 years ago
What is -2.5+ (-4.5)
Zolol [24]

Answer:

-7

Step-by-step explanation:

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8 0
3 years ago
M.
Rasek [7]

Answer: 95.0 square feet

3 0
3 years ago
Creating an Exponential Model
MArishka [77]

Answer:

P = $300

r = 0.15

n = 12

$544.61  (to the nearest cent)

P(1+r)^t

$524.70  (to the nearest cent)

Step-by-step explanation:

P = principal amount = $300

r = annual interest rate in decimal form = 15% = 15/100 = 0.15

n = number of times interest is compounded per unit t = 12

<u>How much she'll owe in 4 years</u>

P = 300

r = 0.15

n = 12

t = 4

P(1+\frac{r}{n})^{nt}=300(1+\frac{0.15}{12})^{12 \times 4}

= $544.61  (to the nearest cent)

<u>Yearly compounding interest rate</u>

<u />P(1+r)^t

<u>How much she'll owe in 4 years at yearly compounding interest</u>

<u />P(1+r)^t=300(1+0.15)^4

= $524.70  (to the nearest cent)

4 0
2 years ago
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