Ask question

Ask question

All categories

3 years ago

8

I need help with #14. I checked on Desmos to see if they represent the same line, which they do. But how do you know if they rep

resent the same line without using Desmos?

1 answer:

DaniilM [7]3 years ago

7 0

Put the second equation so it's in y= form, it should be the same :)

You might be interested in

Solve the inequality.<br> |3x+7| <_ 4

n200080 [17]

Answer:

Using ∣x∣>a⇒x<−a or x>a, we get

∣3x−7∣>4⇒3x−7<−4 or 3x−7>4

⇒3x<3 or 3x>11⇒x<1 or x>

3

11

⇒x∈(−∞,1)∪(

3

11

,∞).

3 0

3 years ago

Read 2 more answers

Simplify 3x + 5x + 14x

navik [9.2K]

Simply collect all the like terms together
So..3+5+14 is 22x

3 0

3 years ago

Read 2 more answers

(x⁴)³<br>ano po answer<br>plss

yanalaym [24]

Answer:

x^12

Step-by-step explanation:

multiply 4 and 3

x^4*3 = x^12

4 0

3 years ago

Write a third-degree polynomial function whose zeros are 1, −3, and 4.

WARRIOR [948]

Recall that zeroes can be transformed into factors by subtracting them from x. This gives us the following factors:

(x - 1)(x + 3)(x - 4)

Now, if you multiply the first two factors together, you get the following:

(x² + 2x - 3)

Multiply that by the last factor, (x - 4), and you get this:

(x³ + 2x² - 3x - 4x² - 8x + 12)

This can be simplified:

(x³ - 2x² - 11x + 12)

And there's your final answer. Hope this helped!

8 0

3 years ago

Read 2 more answers

If N is the population of the colony and t is the time in days, express N as a function of t. Consider Upper N 0 is the origina

lesya [120]

Answer:

(a) $N(t)=Noe^{kt}$

(b)5,832 Mosquitoes

(c)5 days

Step-by-step explanation:

(a)Given an original amount $N_o$ at t=0. The population of the colony with a growth rate $k \neq 0$ , where k is a constant is given as:

$N(t)=Noe^{kt}$

(b)If $N_o=1000$ and the population after 1 day, N(1)=1800

Then, from our model:

N(1)=1800

$1800=1000e^{k}\\$Divide both sides by 1000\\e^{k}=1.8\\$Take the natural logarithm of both sides\\k=ln(1.8)$

Therefore, our model is:

$N(t)=1000e^{t*ln(1.8)}\\N(t)=1000\cdot1.8^t$

In 3 days time

$N(3)=1000\cdot1.8^3=5832$

The population of mosquitoes in 3 days time will be approximately 5832.

(c)If the population N(t)=20,000,we want to determine how many days it takes to attain that value.

From our model

$N(t)=1000\cdot1.8^t\\20000=1000\cdot1.8^t\\$Divide both sides by 1000\\20=1.8^t\\$Convert to logarithm form\\Log_{1.8}20=t\\\frac{Log 20}{Log 1.8}=t\\ t=5.097\approx 5\; days$

In approximately 5 days, the population of mosquitoes will be 20,000.

7 0

3 years ago

Read 2 more answers