Ask question

Ask question

All categories

2 years ago

10

Help me with questions 3 and 4 please!

2 answers:

Vaselesa [24]2 years ago

7 0

Answer:

3.)

y=3X + 1

slope is 3

y intercept is 1

4.)

y=(1/2)X+1

slope is 1/2

y intercept is 1

poizon [28]2 years ago

6 0

Answer:

Question 3: x0 = y1

your basically adding 9 for y!

Question 4: x0 = y1

your adding 1 for y!

Hope this helped!

You might be interested in

Can you pls help me<br> In the diagram,RS and PT are altitudes and m∠Q=68. Find m∠POR

KonstantinChe [14]

Answer:

I believe it is 180

Step-by-step explanation:

All triangles add up to 180

4 0

3 years ago

The height of a soccer ball that is kicked from the ground can be approximated by the function:

kumpel [21]

Why do you want me a little bit of

3 0

3 years ago

if there are more students in one group than the other would it be possible to have equal proportions?

coldgirl [10]

Yes but In the bigger group each person would have less

3 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

6x−3y=5 y−2x=8 How many solutions does the system have?

zhuklara [117]

No Solutions because there are too many = signs.

3 0

3 years ago

Read 2 more answers