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1 year ago

Whats y'alls answer streak, mine is 10. Highest gets brainiest! (not how many questions you have answered.

Mathematics

2 answers:

Nataliya [291]1 year ago

8 0

Answer:

I just started

Step-by-step explanation:

Airida [17]1 year ago

8 0

Answer:

mine was 37 but i broke it 3 days ago :(

so now its 3

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How many groups of 1/2 days are in 1 week? Write a multiplication equation or a division equation to represent the question, the

worty [1.4K]

Answer:

14 groups

Step-by-step explanation:

Given

$Groupings = \frac{1}{2}\ day$

$Duration = 1\ week$

Required

Determine the number of group in the week

To do this, we simply divide the duration by the number half day group.

This gives:

$Group = \frac{Duration}{Groupings}$

$Group = \frac{1\ week}{1/2\ days}$

Convert week to days

$Group = \frac{7\ days}{1/2\ days}$

$Group = \frac{7}{1/2}$

$Group = 7 / \frac{1}{2}$

Convert to multiplication

$Group = 7 * \frac{2}{1}$

$Group = \frac{14}{1}$

$Group = 14$

<em>Hence, there are 14 groups</em>

4 0

2 years ago

Simplify this expression 5a/11a

barxatty [35]

You could cross cancel the a and your final answer would be 5/11

6 0

2 years ago

Read 2 more answers

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

2 years ago

Simplify the expression. <br> −9(−2 − 8) + (−16) −8

Xelga [282]

Answer:

Step-by-step explanation:

Open the parentheses, 18 + 72 - 16 - 8 = 66

7 0

2 years ago

Read 2 more answers

Each basketball comes in a cubic box with side lengths of 1 foot. The locker is 3 feet long, 2 feet wide, and 5 feet high. How m

LuckyWell [14K]

Answer:

30 basketballs

Step-by-step explanation:

one layer would be 3 x 2 = 6

to find how much you would need to cover the bottom of the locker, you would use the numbers 3 and 2, then multiply them and you get 6. You would need 6 basketballs to cover the bottom of the locker. That would be your first layer

5 layers would be 6 x 5 = 30

So the bottom would be filled with 6 basketballs. Now since the locker is 5 feet high multiply 6 x 5 and that would get you your total of basketballs

4 0

2 years ago