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3 years ago

What is the gcf and lcm of 6,12,18

Mathematics

1 answer:

Ierofanga [76]3 years ago

6 0

Answer:

The GCF is 6 and the lcm is 2

Step-by-step explanation:

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Help me explain step by step to each question! please this is due in 6 mins!!

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Answer:

i don't know j dhf jfjyk jrkr htjt jrkr dr rkr

5 0

2 years ago

PLS HELP ,,,

LenKa [72]

The function, g(x), has a constant rate of change and will increase at a faster rate than the function f(x) for all the values of x.

Given:

g(x) = 5/2 x -3 ..... (1)

f(x) = - 3.5 at x = 0

So, putting the value of x=0 in equation (1) for comparison. We get,

g(x) at x = 0

=> g(x) = 5/2 x (0) - 3

=> g(x) = -3

In this value of x function g(x) is faster than function f(x) having a value equal to -3.5.

Similarly, put x = 1 in equation (1) for comparison. We get,

=> g(x) = 5/2 x (1) - 3

=> g(x) = (5-6)/2

=> g(x) = -1/2

In this value of x function g(x) is faster than function f(x) having a value equal to -1.

Similarly, put x = 2 in equation (1) for comparison. We get,

=> g(x) = 5/2 x (2) - 3

=> g(x) = (5-3)

=> g(x) = 2

In this value of x function g(x) is faster than function f(x) having a value equal to 1.5.

Similarly, put x = 3 in equation (1) for comparison. We get,

=> g(x) = 5/2 x (3) - 3

=> g(x) = (15/2 - 3)

=> g(x) = 7.5 - 3

=> g(x) = 4.5

In this value of x function g(x) is faster than function f(x) having a value equal to 4.

Therefore, for all values of x function g(x) is faster than function f(x).

function f(x).

To learn more about the function visit: brainly.com/question/14996787

#SPJ1

8 0

1 year ago

How many hours is this flight: Departure Dallas, TX 8:30 am Arrival Chicago, IL 10:30 am

GenaCL600 [577]

Answer:

$2\ hours$

Step-by-step explanation:

To know the hours of flight, subtract the departure time from the arrival time

$(10:30 am-8:30 am)=2\ hours$

Remember that Chicago and Dallas are in the same time zone

8 0

3 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

3 years ago

Consider the following hypothesis test: H 0: 20 H a: < 20 A sample of 60 provided a sample mean of 19.5. The population stand

viva [34]

Answer:

We reject the null hypothesis and accept the alternate hypothesis. Thus, it be concluded that the population mean is less than 20.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 60

Sample mean, $\bar{x}$ = 19.5