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1 year ago

A number multiplied by 5, subtracted from the sum of 14 and eight times the number

Mathematics

1 answer:

Andrews [41]1 year ago

3 0

Answer:

5m - (14 • 8)

Step-by-step explanation:

pretty sure this is correct

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The table shows the number of books donated to a library each month. Suppose the growth continues exponentially.

telo118 [61]

$\bf \begin{array}{ccll}
\stackrel{t}{months}&\stackrel{A}{books}\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
0&80\\
1&100\\
2&125
\end{array}$ , we know that on Month 0, the Books were 80

$\bf \qquad \textit{Amount for Exponential Growth}\\\\
A=I(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &80\\
I=\textit{initial amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\to &0\\
\end{cases}
\\\\\\
80=I(1+r)^0\implies 80=I\qquad therefore\qquad \boxed{A=80(1+r)^t}$

we also know that on the first month there were 100 books,

$\bf \qquad \textit{Amount for Exponential Growth}\\\\
A=I(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &100\\
I=\textit{initial amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\to &1\\
\end{cases}
\\\\\\
100=80(1+r)^1\implies \cfrac{100}{80}=1+r\implies \cfrac{5}{4}=1+r
\implies 
\cfrac{5}{4}-1=r
\\\\\\
 \cfrac{1}{4}=r\implies 0.25=r\qquad therefore\qquad \boxed{A=80(1+0.25)^t}$

now, how many books when t = 8? A=80(1+0.25)⁸, or A=80(1.25)⁸.

7 0

2 years ago

Read 2 more answers

Ill give brainliest help me out

dusya [7]

Answer:

Step-by-step explanation:

In the graphing calculator, I tried out the different answers, and I think this graph matches yours pretty closely. x is multiplied by a number greater than one, and it looks like instead of expanding/stretching open, it folded in more, which is compressing.

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2 years ago

Christine wants to buy a bicycle for her little brother. She receives a coupon in the mail for 5% off the bicycle. The bicycle h

puteri [66]

The answer is A because you would do 79 x 0.05 which equals 3.95 . So you subtract 3.95 from 79 which is 75.05 minus 50 equal to 25.05

5 0

3 years ago

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6x + -2y² + 4xy² + 3x² + 5xy² simplify it

r-ruslan [8.4K]

$\large\displaystyle\text{$\begin{gathered}\sf 6x+-2y^{2}+4xy^{2}+3x^{2} +5xy^{2} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf Group\:like\:terms \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf =3x^2+6x-2y^2+4xy^2+5xy^2 \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf Sum\:similar\:elements:\:4xy^2+5xy^2=9xy^2 \end{gathered}$}$

$\boxed{\large\displaystyle\text{$\begin{gathered}\sf 3x^2+6x-2y^2+9xy^2 \end{gathered}$} }$

7 0

1 year ago

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for any positive integer n, the sum of the first n positive integers equals n(n 1)/2. what is the sum of all the even integers b

marusya05 [52]

The sum of all the even integers between 99 and 301 is 20200

To find the sum of even integers between 99 and 301, we will use the arithmetic progressions(AP). The even numbers can be considered as an AP with common difference 2.

In this case, the first even integer will be 100 and the last even integer will be 300.

nth term of the AP = first term + (n-1) x common difference

⇒ 300 = 100 + (n-1) x 2

Therefore, n = (200 + 2 )/2 = 101

That is, there are 101 even integers between 99 and 301.

Sum of the 'n' terms in an AP = n/2 ( first term + last term)

= 101/2 (300+100)

= 20200

Thus sum of all the even integers between 99 and 301 = 20200

Learn more about arithmetic progressions at brainly.com/question/24592110

#SPJ4

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1 year ago