Ask question

Ask question

All categories

3 years ago

12

What is the slope of the line ? please help me fast.

2 answers:

Vikki [24]3 years ago

5 0

Answer:1

Step-by-step explanation:

I had to give details

pishuonlain [190]3 years ago

4 0

Answer: -2

Step-by-step explanation: In algebra, we use the word <em>slope</em> to describe how steep a line is and slope can be found using the ratio <em>rise</em> over <em>run</em> between any two points that are on that line.

When finding the slope, first look at y-intercept.

We can see that a point crosses it at 2 units up.

From there, the tie it touches a point

would be to the right 1 on the x-axis.

So the slope is -2/1.

Since we go down 2 and to the right 1 or -2.

You might be interested in

Measure of Angle 1 is 5x.<br> Measure of Angle 3 is x-24.<br><br> What is the measure of Angle 1?

olga55 [171]

Answer:

The answer to your question is: 30°

Step-by-step explanation:

Data

m∠ 1 = 5x

m∠ 3 = x - 24

Process

m ∠ 1 = m ∠ 3 because they are vertical angles

5x = x - 24

5x - x = 24

4x = 24

x = 24 / 4

x = 6

Angle 1 = 5(6)

= 30°

4 0

3 years ago

Use the Distributive Property to solve the equation.

vodka [1.7K]

The answer is the X= -5

6 0

3 years ago

Pipe A can fill a swimming pool in 12 h. Working with another Pipe B, it only takes 3h How long would it take Pipe B working al

MatroZZZ [7]

My guess is 9 hours, because since pipe a takes 12 hours and with pipe b it takes 9, it's 12-9=3 I think.

5 0

3 years ago

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago

Evaluate the numerical expression 2.5(−4.4 − 3.5)

Sophie [7]

<span>2.5(−4.4 − 3.5)
=</span>−4.4 − 3.5=-7.9
=2.5(-7.9)
=-19.75

6 0

3 years ago