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3 years ago

Which expressions are equivalent to

Mathematics

1 answer:

Drupady [299]3 years ago

4 0

Answer:

$\color{Blue}\huge\boxed{Answer}$

D. 16 + 3y +(-3.52)

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The value of a certain car, in dollars, x years from its model year can be predicted by the function f(x)=12,000(0.89)x . The va

scZoUnD [109]

An exponential decay function is:

f=ir^t, f=final amount, i=initial amount, r=common ratio, t=number of times ratio is applied...We can see the initial value is 17000 so all we really have to find is the common ratio, or "rate" as some call it :P If we look at the second point, when t=1 we can see:

14280=17000r

r=357/425

r=0.84

So in five years the SUV will be worth more than the car by:

17000(0.84^5)-12000(0.89^5)

$408.73 (to the nearest cent)

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3 years ago

Read 2 more answers

2 divided by square root 13 plus square root 11

coldgirl [10]

Answer: 3.87132498658 in other words 3.871

Step-by-step explanation:

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2 years ago

There were 32 butterflies in the butterfly exhibit at the zoo. After some new butterflies hatched from their cocoons, there were

BigorU [14]

B and a

hope this helps!

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3 years ago

Read 2 more answers

What is 316+178*2427(2639+26383)-373637/282637

dybincka [34]

Answer:

12537678446.7

Step-by-step explanation:

316+178 divided by 2427 (2639 plus 26383) - 373637/282637

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3 years ago

Which of the following is NOT a linear factor of the polynomial function?

Natalija [7]

Answer:

Among the four choices, $(x + 5)$ is the only one that is not a linear factor of this polynomial function.

Step-by-step explanation:

Let $a$ denote some constant. A linear factor of the form $(x - a)$ is a factor of a polynomial $f(x)$ if and only if $f(a) = 0$ (that is: replacing all $x$ in the polynomial $f(x) \!$ with the constant $a\!$ would give this polynomial a value of $0$ .)

For example, in the second linear factor $(x - 2)$ , the value of the constant is $a = 2$ . Verify that the value of $f(2)$ is indeed $0$ . (In other words, replacing all $x$ in the polynomial $f(x) \!$ with the constant $2$ should give this polynomial a value of $0\!$ .)

$\begin{aligned}f(2) &= 2^3 - 5\times 2^2 - 4 \times 2 + 20 \\ &= 8 - 20 - 8 + 20 \\ &= 0 \end{aligned}$ .

Hence, $(x - 2)$ is indeed a linear factor of polynomial $f(x)$ .

Similarly, it could be verified that $(x - 5)$ and $(x + 2)$ are also linear factors of this polynomial function.

Rewrite the first linear factor $(x + 5)$ in the form $(x - a)$ for some constant $a$ : $(x + 5) = (x - (-5))$ , where $a = -5$ .

Calculate the value of $f(5)$ .

$\begin{aligned}f(5) &= (-5)^3 - 5\times (-5)^2 - 4 \times (-5) + 20 \\ &= (-125) - 125 + 20 + 20 \\ &= -210\end{aligned}$ .

$f(5) \ne 0$ implies that $(x - (-5))$ (which is equivalent to $(x + 5)$ ) isn't a linear factor of this polynomial function.

3 0

3 years ago