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3 years ago

Write 14 hundred thousands, 19 ten thousands, 17 thousands, 7 hundreds 15 ones

Mathematics

1 answer:

Kryger [21]3 years ago

6 0

1,607,715 Is the answer I hope that helps

plz mark brainliest

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Translate the sentence into an equation. Five more than the product of a number and 9 is equal to 6

kap26 [50]

The first thing we will do for this case is to define the following variable:
x = real number.
We then have the following sentence:
"Five more than the product of a number and 9 is equal to 6"
The equation sought is:
9x + 5 = 6
Answer:
An equation for the sentence is:
9x + 5 = 6

6 0

3 years ago

each hour the number of bacteria in dr. nall's petri dish tripled what percent is the population compared to the population the

yan [13]

The starting percent is 100%. After one hour, bacteria tripled, so it is 300%.

3 0

2 years ago

55) A shirt that usually costs $30 is on sale for $20. Find the percent of<br> the discount.

ivanzaharov [21]

Find the amount of the discount and divide by the original price:

30-20 = 10

10/30 = 0. 333

Multiply by 100 for percent:

0.333 x 100 = 33.3333%

Round the answer as needed.

4 0

2 years ago

Read 2 more answers

An Impressionist painting increases in value

jok3333 [9.3K]

Answer:

A = $8406.6

Step-by-step explanation:

Given:

Average rate $r=9\%$

Initial cost of painting $a = \$1500$

Time $t = 20\ years$

We need to find the final amount of painting at the end of a 20-year.

Solution:

Using Exponential Growth rate formula as:

$A = a(1+r)^t$ ----------(1)

Where:

A = Final amount

a = Initial amount.

r = Rate as a decimal.

t = Time.

Now, we substitute all given values in equation 1.

$A = 1500(1+0.09)^{20}$

$A = 1500(1.09)^{20}$

Substitute $1.09^{20} = 5.60$ in above equation.

$A = 1500\times 5.60$

A = $8406.62

Therefore, value of the painting at the end of a 20-year A = $8406.6

5 0

3 years ago

Suppose John opens a savings account with $1,000 that compounds interest daily. The APR at the time John deposits the account is

Fed [463]

$\bf \qquad \qquad \textit{Annual Yield Formula}
\\\\
\left. \qquad \qquad \right.\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 3.5\%\to \frac{3.5}{100}\to &0.035\\
n=
\begin{array}{llll}
\textit{times it compounds per year}
\end{array}\to &365
\end{cases}
\\\\\\
\left(1+\frac{0.035}{365}\right)^{365}-1$

4 0

3 years ago