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zlopas [31]
2 years ago
12

What is the table for the rule y-2x+7

Mathematics
1 answer:
Mandarinka [93]2 years ago
5 0
As an example I will use 3 for x. First you would do 3 times 2 which is 6 and then 6 plus 7 is 13 and then you answer is y-16. 
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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Nady [450]

Answer:

A) 2x³+11x²+8x-16

Step-by-step explanation:

When you multiply s(x) by t(x) you get something like this:

s(x) \times t(x) = (2 {x}^{2}  + 3x - 4) \times (x + 4) \\  = 2 {x}^{3}  + 3 {x}^{2}  - 4x + 8 {x}^{2}  + 12x - 16 \\  = 2 {x}^{3}  + 11 {x}^{2}  + 8x - 16

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According to the table, what is the relative
krok68 [10]

Answer:

About 67%

Step-by-step explanation:

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3 0
2 years ago
¿Se pueden aplicar las razones trigonométricas de ángulos fijos para resolver cualquier tipo de triangulo?
DerKrebs [107]

Answer:

Solo se puede usarlas con triángulos rectángulos (right triangles en ingles).

5 0
3 years ago
Sally Sue had spent all day preparing for the prom. All the glitz and the glamour of the evening fell apart as she stepped out o
Nuetrik [128]

We have the function

p(t)=550(1-e^{-0.039t})

Therefore we want to determine when we have

p(t_0)=550

It means that the term

e^{-0.039t}

Must go to zero, then let's forget the rest of the function for a sec and focus only on this term

e^{-0.039t}\rightarrow0

But for which value of t? When we have a decreasing exponential, it's interesting to input values that are multiples of the exponential coefficient, if we have 0.039 in the exponential, let's define that

\alpha=\frac{1}{0.039}

The inverse of the number, but why do that? look what happens when we do t = α

e^{-0.039t}\Rightarrow e^{-0.039\alpha}\Rightarrow e^{-1}=\frac{1}{e}

And when t = 2α

e^{-0.039t}\Rightarrow e^{-0.039\cdot2\alpha}\Rightarrow e^{-2}=\frac{1}{e^2}

We can write it in terms of e only.

And we can find for which value of α we have a small value that satisfies

e^{-0.039t}\approx0

Only using powers of e

Let's write some inverse powers of e:

\begin{gathered} \frac{1}{e}=0.368 \\  \\ \frac{1}{e^2}=0.135 \\  \\ \frac{1}{e^3}=0.05 \\  \\ \frac{1}{e^4}=0.02 \\  \\ \frac{1}{e^5}=0.006 \end{gathered}

See that at t = 5α we have a small value already, then if we input p(5α) we can get

\begin{gathered} p(5\alpha)=550(1-e^{-0.039\cdot5\alpha}) \\  \\ p(5\alpha)=550(1-0.006) \\  \\ p(5\alpha)=550(1-0.006) \\  \\ p(5\alpha)=550\cdot0.994 \\  \\ p(5\alpha)\approx547 \end{gathered}

That's already very close to 550, if we want a better approximation we can use t = 8α, which will result in 549.81, which is basically 550.

Therefore, we can use t = 5α and say that 3 people are not important for our case, and say that it's basically 550, or use t = 8α and get a very close value.

In both cases, the decimal answers would be

\begin{gathered} 5\alpha=\frac{5}{0.039}=128.2\text{ minutes (good approx)} \\  \\ 8\alpha=\frac{8}{0.039}=205.13\text{ minutes (even better approx)} \end{gathered}

7 0
1 year ago
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