All categories

3 years ago

If g(x) = - 2 x^2 + 6 , find g(x) when x= - 4 Pls answer

Mathematics

1 answer:

laila [671]3 years ago

7 0

Answer:

PLZ MARK ME BRAINLIEST

You might be interested in

8 to the second power to the -4 power is equivalent to

jeyben [28]

Answer:

1.138789 is equivalent to 8 to the second power to the -4 power.

Hope this helps :)

3 0

2 years ago

A piece of chalk is 34/56 centimeter long. Brian breaks off 23/56 centimeter long. Sara added 12/56 centimeter to the chalk that

mr Goodwill [35]

Answer:

23/56 centimeters

Step-by-step explanation:

all of the variables have a common denominator, so you can effectively ignore the denominator when solving.

1. you start off with 34/56 centimeter of chalk and Brian breaks off 23/56 centimeters worth, leaving you with 11/56 centimeters

$\frac{34}{56} -\frac{23}{56} =\frac{11}{56}$

2. Sara then added 12/56 centimeters worth, making the final amount 23/56 centimeters of chalk

$\frac{11}{56}+\frac{12}{56} =\frac{23}{56}$

An equation to summarize all of this is:

$\frac{34}{56} -\frac{23}{56} +\frac{12}{56}$

7 0

4 years ago

How to determine how many cubes fit inside a rectangular prism?

Sedbober [7]

If you're talking about unit cubes, then the equation is

length*width*height

5 0

3 years ago

Who wanna give me the answer ;)

patriot [66]

Answer:

Step-by-step explanation:

6 0

3 years ago

How much more would $1000 earn in 5 years in an account compounded continuously than an account compounded quarterly if the inte

choli [55]

$\bf ~~~~~~ \textit{Compounding Continuously Interest Earned Amount}\\\\
A=Pe^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to& \$1000\\
r=rate\to 3.7\%\to \frac{3.7}{100}\to &0.037\\
t=years\to &5
\end{cases}
\\\\\\
A=1000e^{0.037\cdot 5}\implies A=1000e^{0.185}\\\\
-------------------------------\\\\$

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3.7\%\to \frac{3.7}{100}\to &0.037\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, thus four}
\end{array}\to &4\\
t=years\to &5
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.037}{4}\right)^{4\cdot 5}\implies A=1000(1.00925)^{20}$

compare both amounts.

8 0

3 years ago