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2 years ago

Urgent.

Mathematics

1 answer:

den301095 [7]2 years ago

6 0

Answer:

#1. 105.5 because of the nine 4 goes to 5

#2. 67,990,000 because of the 4 it doesn't go up

#3. -4.1 because of the 4 the 0 goes to 1

if the number place before the one you are rounding is 5,6,7,8,or 9you go up by one and the rest becomes zeros

if the number place before the one you are rounding is 0,1,2,3,or 4 you stay the same and the rest becomes zeros

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Need answers for this homework. plz help due to tomorrow

KIM [24]

2a) 30
2 x 15
3 x 5
30=2x3x5
2b) 60
6 x 10
2x3 2x5
60=2x3x2x5

4 0

3 years ago

What is the period of the sinusoidal function? Enter your answer in the box.

muminat

Answer:

ANSWER:

The period of the sinusoidal function is 2π.

Step-by-step explanation:

EXPLANATION:

The sinusoidal function is a periodic function.It goes one unit up and one unit down with an amplitude of one and it repeats itself after this time interval.So,the period of the sine function is 2π

A sine or sinusoidal loop is a perpetual swing. It is called after the role sine. It happens frequently in tentative and practiced math, science, physics, engineering, signal processing, and other disciplines. Its utmost fundamental pattern as a role of time (t).

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3 years ago

If bc is the diameter of the circle below, what is the area of the circle

dangina [55]

Answer:

option a.

radius of the circle=3 units.

area of the circle= pi×r²

area of circle=pi×3²

area of circle=pi×9units²

5 0

3 years ago

Read 2 more answers

HELP<br> If you don’t know it then please don’t answer.

vovangra [49]

the answer is 4.8. Your welcome

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