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2 years ago

What is an equation of the line that passes through the points (-4, 2) and (-8, -3)?

1 answer:

4 0

To find this use slope (m) for y2-y1/x2-x1
The find the y intercept by using the formula y-y1= m(x-x1)
Using these formulas the answer is
Y=1.25 or 5/4x+5

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Help pls ill give extra points

Nadya [2.5K]

Suzuki after line 3 where it says 42 / 5 +16 she shouldn’t have added 5 and 16 she needs to follow pemdas and do 42/5 which is 8.4 then she needs to add 8.4 and 16 to get an answer of 24.4

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2 years ago

Plot a point on the coordinate plane to represent each of the

Juli2301 [7.4K]

(1,4) you go right 1 time, and than go up 4.

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2 years ago

A boat is pulled toward a dock by means of a rope wound on a drum that is located 6 ft above the bow of the boat. If the rope is

Monica [59]

Answer:

Step-by-step explanation:

Given that:

The height of the dock (h) = 6

Let represent d to be the distance between the boat and the dock

Let the length of the rope between the boat and the drum be denoted by (l)

Then, the rate of change for the length of the rope be:

dl/dt = -5 ft/s

Using Pythagoras rule to determine the relationship between these values, we have:

$l^2 = h^2 +d^2$

$l^2 = 6^2 + d^2$

$l^2 = 36 + d^2$

We relate to: $2l * \dfrac{dl}{dt} = 2d* \dfrac{dd}{dt}$

From the question;

l = 34,

So to find $\dfrac{dd}{dt}$ , we get;

$d = \sqrt{l^2 - 36}$

$d = \sqrt{34^2 - 36}$

$d = \sqrt{1156- 36}$

$d = \sqrt{1120}$

d = 33.46

So, we have:

$\dfrac{dd}{dt}= \dfrac{l}{d} \times \dfrac{dl}{dt}$

$\dfrac{dd}{dt}= \dfrac{34}{33.46} \times - 5$

$\dfrac{dd}{dt}= 1.016 \times - 5$

$\dfrac{dd}{dt}=-5.08 \ ft/sec$

4 0

2 years ago

Write a function for the situation discribed and find the value after 5 yrs. A $12,500 car depreciates 9% each year. Please help

sweet [91]

$\bf \qquad \textit{Amount for Exponential Decay}\\\\
A=P(1 - r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{initial amount}\to &12500\\
r=rate\to 9\%\to \frac{9}{100}\to &0.09\\
t=\textit{elapsed time}\\
\end{cases}
\\\\\\
A=12500(1-0.09)^t\implies A=12500(0.91)^t
\\\\\\
\textit{after 5 years t = 5}\qquad A=12500(0.91)^5$

and surely you know how much that is.

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3 years ago

1 + 200000 - 1 + 200000

Valentin [98]

1+200000-1+200000
200001-1+200000
200000+200000
400000

4 0

2 years ago