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3 years ago

#8 I need help with the y-intercept

Mathematics

1 answer:

guajiro [1.7K]3 years ago

5 0

The y intercept is 2

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Please select the word from the list that best fits the definition

lana66690 [7]

Answer:

There is nothing there...

Step-by-step explanation:

There is nothing at all.

6 0

3 years ago

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Value of peters house has risen 20% a year for the last three years. If the value of peters home three years ago was $120 thousa

notka56 [123]

Answer:

The value now = $207,360

Step-by-step explanation:

Rate of increase = 20% per year = 20/100 = 0.2 per year

value 3 years ago = $120,000

interest at year 1:

20% of 120,000

= 0.2 × 120,000 = 24,000

Value at the end of year 1

= initial value + interest

= 120,000 + 24,000 = $144,000

In year 2:

20% of 144,000

= 0.2 × 144,000 = 28,800

Value at the end of year 2:

144,000 + 28,800 = $172,800

In year 3

0.2 × 172,800 = 34,560

Value at the end of year 3:

34,560 + 172,800 = $207,360

8 0

3 years ago

What number is being factored in this factor tree? A.72 B.180 C.216 D.324

Tatiana [17]

Answer:

not sure if it's ryt but I think it's A

7 0

3 years ago

A species of beetles grows 32% every year. Suppose 100 beetles are released into a field. How many beetles will there be in 10 y

siniylev [52]

Given that a species of beetles grows 32% every year.

So growth rate is given by

r=32%= 0.32

Given that 100 beetles are released into a field.

So that means initial number of beetles P=100

Now we have to find about how many beetles will there be in 10 years.

To find that we need to setup growth formula which is given by

$A=P(1+r)^n$ where A is number of beetles at any year n.

Plug the given values into above formula we get:

$A=100(1+0.32)^n$

$A=100(1.32)^n$

now plug n=10 years

$A=100(1.32)^{10}=100(16.0597696605)=1605.97696605$

Hence answer is approx 1606 beetles will be there after 20 years.

Now we have to find about how many beetles will there be in 20 years.

To find that we plug n=20 years

$A=100(1.32)^{20}=100(257.916201549)=25791.6201549$

Hence answer is approx 25791 beetles will be there after 20 years.

Now we have to find time for 100000 beetles so plug A=100000

$A=100(1.32)^n$

$100000=100(1.32)^n$

$1000=(1.32)^n$

$log(1000)=n*log(1.32)$

$\frac{\log\left(10000\right)}{\log\left(1.32\right)}=n$

33.174666862=n

Hence answer is approx 33 years.

4 0

3 years ago

Read 2 more answers

Plss help I don’t understand it it’s for homework

tia_tia [17]

Answer: r= -3

Next 3 terms following 54 are -162, 486, -1458

Step-by-step explanation:

7 0

3 years ago