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2 years ago

Would you please help me with this quiz answer, I will really appreciate if you do.

Mathematics

1 answer:

laila [671]2 years ago

6 0

Answer:

a . domain 5,0,7,9,0

range -2,-2,-4,8,2

b. domain 2,4,8,9

range 1,2,4,11

Step-by-step explanation:

<h3>a is not a function</h3>

because function is a relationship in which each domain element occurs only once.

<h3>b is a function</h3>

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A city has a population of 230,000 people. Suppose that each year the population grows by 9%. What will the population be after

zloy xaker [14]

<u>Answer:</u>

Step-by-step explanation:

If it grows by 9% each year, you can simply do the current population times 109%. Because its an additional 9%.

230,000*109% = 250,700

After the first year, there is a population of 250,700 people.

250,700*109% = 273,263

After the second year, there is a population of 273,263 people.

273,263*109% = 297,856.67

After the third year, there is a population of 297,856.67 people.

297,856.67*109% = 324,663.77

After the fourth year, there is a population of 324,663.77 people.

324,663.77*109% = 353,883.509

After the fifth year, there is a population of 353,883.509 people.

353,883.509*109% = 385,733.025

After the sixth year, there is a population of 385,733.025 people

Round it up to 385,733 people.

7 0

2 years ago

Factor 18x+6y using gcf

Andrew [12]

Group and factor out the greatest common factor then combine. 6(3x+y)

5 0

3 years ago

Factor the expression completely. 9x+6

Kamila [148]

Answer:

3(3x+2)

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

In which equation does x = 15?

S_A_V [24]

To help you with it, plug-in 15 in each equation and see what you get!

8 0

2 years ago

Read 2 more answers

Use the exponential decay model, Upper A equals Upper A 0 e Superscript kt, to solve the following. The half-life of a certai

Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

$A=A_0 e^{kt}$

Where,

$A_0$ = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = $\frac{A_0}{2}$ ,

i.e.

$\frac{A_0}{2}=A_0 e^{19k}$

$\frac{1}{2} = e^{19k}$

$0.5 = e^{19k}$

Taking ln both sides,

$\ln(0.5) = \ln(e^{19k})$

$\ln(0.5) = 19k$

$\implies k = \frac{\ln(0.5)}{19}\approx -0.03648$

Now, if the substance to decay to 78% of its original amount,

Then $A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0$

$0.78 A_0=A_0 e^{-0.03648t}$

$0.78 = e^{-0.03648t}$

Again taking ln both sides,

$\ln(0.78) = -0.03648t$

$-0.24846=-0.03648t$

$\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7$

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0

2 years ago

Would you please help me with this quiz answer, I will really appreciate if you do. ​

Would you please help me with this quiz answer, I will really appreciate if you do.