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3 years ago

Find the slope a line that passes between (-5, 8) and (2,4)

Mathematics

1 answer:

sveta [45]3 years ago

6 0

Answer:

Step-by-step explanation:

formula : m=(y2-y1)/(x2-x1)

(4 - 8) / (2 - - 5)

-4 / 7

m = -4 / 7 = -0.57143

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Order the following numbers from least to greatest -9/2 0 1 3/5 9/7 2/3

marysya [2.9K]

$\frac{-9}{2}$

$\frac{2}{3}$

$\frac{9}{7}$

1 $\frac{3}{5}$

3 0

3 years ago

Read 2 more answers

A chemical substance has a decay rate of 6.9% per day. The rate of change of an amount N of the chemical is given by the equatio

GaryK [48]

Answer:

a) $N=N_0e^{-0.069t}$

b) $N=696.9$ grams

c) $t=10$ days

Step-by-step explanation:

We are going to use separation of variables to solve.

Get all your t's to one side and your N's to opposing side.

$\frac{dN}{dt}=-0.069N$

Multiply both sides by $dt$ :

$dN=-0.069N dt$

Divided both sides by $N$ :

$\frac{dN}{N}=-0.069 dt$

Integrate both sides:

$\ln|N|=-0.069t+C$

The equivalent exponential form is:

$e^{-0.069t+C}=N$

Using law of exponents you can write this as:

$e^{-0.069t}e^C=N$

$e^C$ is just a positive constant that I'm going to replace with K:

$e^{-0.069t}K=N$

Applying the symmetric property of equality:

$N=e^{-0.069t}K$

Applying the commutative property of multiplication:

$N=Ke^{-0.069t}$

K actually represents the initial amount of chemical substance since when plugging in 0 for t you get K for N, like so:

$N=Ke^{-0.069 \cdot 0}$

$N=Ke^{0}$

$N=K(1)$

$N=K$

We are given at time 0 the amount of chemical substance,N, is K. They want us to represent this value with $N_0$ instead. So the exponential equation is:

$N=N_0e^{-0.069t}$

We are given $N_0=800$ at $t=0$ .

We are asked to find how much of the chemical substance, N, remains after 2 days. So we replace t with 2 in $N=800e^{-0.069t}$ :

$N=800e^{-0.069 \cdot 2}$

Put into calculator:

$N=696.9$ (this was rounded to the nearest tenths)

The last part is asking for how many days will it take a initial 800 grams to go down to half of 800 grams.

We need to see the following equation:

$\frac{1}{2}(800)=800e^{-0.069t}$

$400=800e^{-0.069t}$

Divide both sides by 800:

$\frac{400}{800}=e^{-0.069t}$

Reduce the fraction:

$\frac{1}{2}=e^{-0.069t}$

Convert to logarithmic form:

$\ln(\frac{1}{2})=-0.069t$

Divide both sides by -0.069:

$\frac{\ln(\frac{1}{2})}{-0.069}=t$

Input into calculator:

$10.0=t$

$t=10.0$

$t=10$

6 0

4 years ago

If f(x) = 6x – 9, evaluate for x = -5.

Katarina [22]

Answer:

6(-5)-9

-30-9

=-39

Step-by-step explanation:

3 0

3 years ago

Lilianna uses 3/4 calories per minute just by sitting. She uses 1 more calorie per minute by walking. Liliana uses a total of 12

Anon25 [30]

Answer:

Variable d represents the time per minute Lilianna spends in the park.

Step-by-step explanation:

Lilianna uses $\frac{3}{4}$ calories per minute just by sitting. She uses 1 more calorie per minute by walking.

Liliana uses a total of $12\frac{1}{4}$ calories walking to the park.

Lilianna uses the equation, $d(\frac{3}{4}+1)=12\frac{1}{4}$ to represent the situation.

Here, variable d represents the time per minute she spends in the park.

and is given as,

$d(\frac{3}{4}+1)=12\frac{1}{4}$

It can be written as,

$d(\frac{3}{4}+1)=\frac{49}{4}$

Solving for d,

$d(\frac{3}{4}+1)=\frac{49}{4}$

$\Rightarrow d(\frac{3+4}{4})=\frac{49}{4}$

$\Rightarrow d(\frac{7}{4})=\frac{49}{4}$

$\Rightarrow d=\frac{49 \times 4}{4 \times 7}$

$\Rightarrow d=7$

Thus, she spend d= 7 minutes total to the park.

5 0

3 years ago

How far above sea level is the ranger station?

Schach [20]

Since he descended 12 meters, we subtract this from the overall height of Mount Ka'ala, so then we are only calculating how high ABOVE the sea level it is.

1232 - 12 = 1220

The height of Mount Ka'ala is therefore 1,220 meters.
To calculate how much a fifth of Mount Ka'ala is (since the ranger station is 2/5's up), we would divide this number by 5

1220 ÷ 5 = 244

Since ONE fifth of the height is 244 meters, TWO fifths would be double that amount.

244 x 2 = 488

488 meters.
The ranger station is 488m above sea level.

3 0

3 years ago