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3 years ago

Will give brainliest to first correct answer!! No links!! Help!!

Mathematics

2 answers:

andriy [413]3 years ago

8 0

Answer:

the top box plot should be the way to go

Lelechka [254]3 years ago

5 0

Answer:

this is the answer

Step-by-step explanation:

I hope it was helpful

B is the answer

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PLEASE HELP this is a formula for a sequence, f(x) = 8 - 11x, what is f(8)?

Elenna [48]

Answer:

-80

Step-by-step explanation:

plug in 8 to the equation

8 - 11 (8)

multiply 8 and 11 first: 8 - 88

add 8 to -88: -80

7 0

2 years ago

Mary spent a total of $353.86 for a party. She spent $200.51 on food, plus an additional $30.67 for each hour of the party. How

BartSMP [9]

The party was 5 hours. 353.86-200.51=153.35 then you divide that number by 30.67 and you get 5.

5 0

2 years ago

How does this work?????

ankoles [38]

Answer:

x = 4 $\sqrt{5}$

Step-by-step explanation:

Since the triangle is right, solve for x using Pythagoras' identity, that is

x² = 8² + 4² = 64 + 16 = 80

Take the square root of both sides

x = $\sqrt{80}$ = $\sqrt{16(5)}$ = $\sqrt{16}$ × $\sqrt{5}$ = 4 $\sqrt{5}$ ← exact value

3 0

3 years ago

You are saving money to buy an electric guitar. You deposit $1000 in an account that earns interest compounded annually. The exp

Katena32 [7]

Let's move like a crab, backwards some.

after 2 years?

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &2
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 2}\implies A=1000(1.03)^2$

after 3 years?

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &3
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\implies A=1000(1.03)^3$

is that enough to pay the $1100?

now, let's write 1000(1+r)² in standard form

1000( 1² + 2r + r²)

1000(1 + 2r + r²)

1000 + 2000r + 1000r²

1000r² + 2000r + 1000 <---- standard form.

8 0

3 years ago

5. What do you notice about the numbers in the last two columns of the table? Answer?

Lostsunrise [7]

Photo????? i need more context:)

8 0

3 years ago