All categories

3 years ago

What is y=3(2x+1)+5 work need to be shown

Mathematics

1 answer:

jonny [76]3 years ago

6 0

Answer:

In slope intercept form, this equation is y = 6x + 8

Step-by-step explanation:

All you need to do to get this to slope intercept form is to distribute the term and then combine like terms.

y = 3(2x + 1) + 5 ----> Distribute

y = 6x + 3 + 5 ----> Combine like terms

y = 6x + 8

You might be interested in

Suppose that $9500 is placed in an account that pays 8% interest compounded each year.

Blababa [14]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$9500\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{each year, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1 \end{cases}$

$\bf A=9500\left(1+\frac{0.08}{1}\right)^{1\cdot 1}\implies A=9500(1.08)\implies \boxed{A=10260} \\\\\\ \stackrel{\textit{after 2 years, t = 2}}{A=9500\left(1+\frac{0.08}{1}\right)^{1\cdot 2}}\implies A=9500(1.08)^2\implies \boxed{A=11080.8}$

6 0

3 years ago

What is a mathematical statement that two expressions are equal

natulia [17]

It is called an equation.

6 0

2 years ago

Learning Thoery In a learning theory project, the proportion P of correct responses after n trials can be modeled by p = 0.83/(1

elena-s [515]

Answer:

a) $P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536$

b) $P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666$

c) $0.75 =\frac{0.83}{1+e^{-0.2n}}$

$1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}$

$e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}$

$ln e^{-0.2n} = ln (\frac{8}{75})$

$-0.2 n = ln(\frac{8}{75})$

And then if we solve for t we got:

$n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials$

d) If we find the limit when n tend to infinity for the function we have this:

$lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83$

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

Step-by-step explanation:

For this case we have the following expression for the proportion of correct responses after n trials:

$P(n) = \frac{0.83}{1+e^{-0.2t}}$

Part a

For this case we just need to replace the value of n=3 in order to see what we got:

$P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536$

So the number of correct reponses after 3 trials is approximately 0.536.

Part b

For this case we just need to replace the value of n=7 in order to see what we got:

$P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666$

So the number of correct responses after 7 weeks is approximately 0.666.

Part c

For this case we want to solve the following equation:

$0.75 =\frac{0.83}{1+e^{-0.2n}}$

And we can rewrite this expression like this:

$1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}$

$e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}$

Now we can apply natural log on both sides and we got:

$ln e^{-0.2n} = ln (\frac{8}{75})$

$-0.2 n = ln(\frac{8}{75})$

And then if we solve for t we got:

$n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials$

And we can see this on the plot attached.

Part d

If we find the limit when n tend to infinity for the function we have this:

$lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83$

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

5 0

3 years ago

Write all integers between: a. -17 and -12 b. -1 and -6

egoroff_w [7]

Answer:

look at explanation

Step-by-step explanation:

a) -13,-14,-15,-16

b) -2,-3,-4,-5

5 0

2 years ago

Margie walked 3.75 miles on Tuesday morning. On Wednesday morning, she walked 1.2 times as far as she did on Tuesday. How far di

kari74 [83]

She walked 4.5 miles. just multiply the miles she walked Tuesday by 1.2.

5 0

2 years ago