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1 year ago

Find the point-slope equation for

1 answer:

4 0

$(\stackrel{x_1}{7}~,~\stackrel{y_1}{-21})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{23}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{23}-\stackrel{y1}{(-21)}}}{\underset{run} {\underset{x_2}{-4}-\underset{x_1}{7}}} \implies \cfrac{23 +21}{-4 -7} \implies \cfrac{ 44 }{ -11 }\implies -4$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-21)}=\stackrel{m}{-4}(x-\stackrel{x_1}{7})\implies y+21=-4(x-7)$

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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.

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3 years ago

A fish was swimming LaTeX: 2\frac{1}{2}2 1 2 feet below the water's surface at 1:00 p.M. Three hours later, the fish was at a de

marin [14]

Answer:

$\frac{23}{4}$ feet

Step-by-step explanation:

The depth of the fish to the surface of water at 1:00 pm = 2 $\frac{1}{2}$ feet

= $\frac{5}{2}$ feet

The depth of the fish to its initial position at 4:00 pm = 3 $\frac{1}{4}$ feet

= $\frac{13}{4}$ feet

The position of the fish with respect to the water's surface at 4:00 pm = the depth of the fish to the surface of water at 1:00 pm + the depth of the fish to its initial position at 4:00 pm

= $\frac{5}{2}$ + $\frac{13}{4}$

= $\frac{10 + 13}{4}$

= $\frac{23}{4}$

The position of the fish with respect to the water's surface at 4:00 pm is $\frac{23}{4}$ feet.

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3 years ago

What is 6.338 rounded to the nearest hundredth

azamat

Answer:

6.34

Step-by-step explanation:

brainliest pleassse

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3 years ago

Solve the equation ?

Morgarella [4.7K]

6m+1=-23. Do the inverse operation of adding 1 which is to subtract 1 to both sides =====> 6m+1-1= -23-1

You divide by 6 both sides
6m/6=-24/6

m= -4

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3 years ago

Which expression is equivalent to 4x+6y-8y?

lutik1710 [3]

Answer:

4x-2y

Step-by-step explanation:

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3 years ago

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