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2 years ago

Kenny wrote the equation for a linear relationship shown below

Mathematics

2 answers:

Alex777 [14]2 years ago

5 0

Y is equal to -17. This is the answer becuase -3 times 7 is -21 and -21 plus 4 is equal to -17.

Ilia_Sergeevich [38]2 years ago

4 0

Y would equal -17! Good luck!

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A club has 12 members. In how many ways can we select four members to go on a trip?

ivanzaharov [21]

This is the number of combinations of 4 from 12

12C4 = 12!
----------
4! 8!

= 12*11*10*9 11*5*9 = 495
-------------- =
4*3*2*1

3 0

3 years ago

Which expression is equivalent to 4/9x7?

LiRa [457]

Is there an answer key?

8 0

2 years ago

Solve x − 5y = 6 for x.

pentagon [3]

Answer:

All real numbers

Step-by-step explanation:

Solve a system of equations by elimination

With the equation given, you can solve for y to get: y = $\frac{1}{5}$ x - $\frac{6}{5}$

Then, substitute this value in on the equation.

x - 5( $\frac{1}{5}$ x - $\frac{6}{5}$ ) = 6

x - x + 6 = 6

6 = 6

The solution is all real numbers

8 0

3 years ago

A rectangle garden is 6 feet long and 4 feet wide a second rectangular garden has dimensions that are doubled the dimensions of

OLga [1]

I do not know this one but the first garden is 24 and the second one is 576.

8 0

2 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