All categories

2 years ago

Please answer the question 3x < 18

Mathematics

1 answer:

Ann [662]2 years ago

4 0

3x<18 I'm not sure what the rule for this problem is.

You might be interested in

-6/5n=8 please help me

valentina_108 [34]

N=-3/20
In decimal form it would be n=-0.15

8 0

2 years ago

I need a lot of help please! ):<br><br> If you don’t know please do not comment.

damaskus [11]

Step-by-step explanation:

scale up :

.05

1/2

scale down :

102

7 0

2 years ago

Read 2 more answers

Express 80 as the product of its prime factors.<br> Write the prime factors in ascending order.

stepan [7]

Answer:

the answer of that is 16 x 5

Step-by-step explanation:

7 0

1 year ago

22 students is __% of 50 is 10

nekit [7.7K]

Answer: it might be 0.22 or 0.50

Step-by-step explanation:

cause i did 22/100 and 50/100 and got these answer one of them might be right or both of them are wrong. This is what i think so.

6 0

2 years ago

Derek found a function that approximately models the population of iguanas in a reptile garden, where x represents the number of

serious [3.7K]

Answer:

$i(x)=12 \times (1+\frac{0.9}{12})^{12x}$ and growth rate factor is 0.075

Step-by-step explanation:

The function that models the population of iguanas in a reptile garden is given by $i(x)=12 \times (1.9)^{x}$ , where x is the number of years.

Since, $i(x)=12 \times (1.9)^{x}$

i.e. $i(x)=12 \times (1+0.9)^{x}$ .

Therefore, the monthly growth rate function becomes,

i.e. $i(x)=12 \times (1+\frac{0.9}{12})^{x \times 12}$ .

i.e. $i(x)=12 \times (1+\frac{0.9}{12})^{12x}$ .

Hence, the monthly growth rate is i.e. $i(x)=12 \times (1+\frac{0.9}{12})^{12x}$ .

Also, the growth factor is given by $\frac{0.9}{12}$ = 0.075.

Thus, the growth factor to nearest thousandth place is 0.075.

4 0

2 years ago