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3 years ago

14

at the theater,the worth family spent $18.00 on adult tickets, $16.50 on children`stickets,and $11.75 on the refreshments. How m

uch did they spend in all.

2 answers:

Goryan [66]3 years ago

7 0

They spent a total of $46.25

Here's how I got my answer:

Step 1: Add $18.00 and $16.40 together. Which is $34.50.

Step 2: Add $34.50 and $11.75, which will give you your answer, $46.25.

horrorfan [7]3 years ago

4 0

$46.25 was spent total.

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ladessa [460]

Answer:

B. $1.83\text{ m}^2$

Step-by-step explanation:

We are asked to find the body surface area of a person whose height is 150 cm and who weighs 80 kg.

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Substitute the values:

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3 years ago

Read 2 more answers

Describe the end behavior of a ninth-degree polynomial with a negative leading coefficient

Degger [83]

Given:

The degree of polynomial = 9

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To find:

The end behavior of the polynomial.

Solution:

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If the degree of a polynomial is odd and leading coefficient is negative, then

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3 years ago

This Question: 1 pt

katrin2010 [14]

Answer:

No Solution.

Step-by-step explanation:

x+y=8

y=3-x

-------------

x+(3-x)=8

x+3-x=8

x-x+3=8

0+3=8

3=8

no solution

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3 years ago

Find the value of $ 15,000 at the end of one year if it is invested in an account that has an interest rate of 4.95 % and is com

lora16 [44]

A)

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$15000\\
r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{twelve months, thus}
\end{array}\to &12\\
t=years\to &1
\end{cases}
\\\\\\
A=15000\left(1+\frac{0.0495}{12}\right)^{12\cdot 1}$

b)

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$15000\\
r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{365 days, thus}
\end{array}\to &365\\
t=years\to &1
\end{cases}
\\\\\\
A=15000\left(1+\frac{0.0495}{365}\right)^{365\cdot 1}$

c)

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$15000\\
r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{four quarters, thus}
\end{array}\to &4\\
t=years\to &1
\end{cases}
\\\\\\
A=15000\left(1+\frac{0.0495}{4}\right)^{4\cdot 1}$

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