How many multiples of 9 are there between 100 and 2,000?

Mathematics

2 answers:

Marianna [84]3 years ago

8 0

Answer: First Option

The number of multiples of 9 is 211

Step-by-step explanation:

Note that the first multiplo of 9 between 100 and 2000 is number 108. The last multiple of 9 is 1998.

If we use the arithmetic sequence to perform the calculation

$a_n = a_1 + d(n - 1)$

So the first term $a_1$ is:

$a_1 = 108$

The last term $a_n$ is

$a_n = 1998$

The common difference is

$d = 9$

Thus

$1998 = 108 + 9(n-1)$

We solve the equation for n and obtain the number of multiples of 9.

$1998-108 = 9(n-1)\\\\\frac{1998-108}{9}=n-1\\\\n = 210 +1\\\\n = 211$

Nookie1986 [14]3 years ago

8 0

196.00000000000000000000000

You might be interested in

The flagpole is 20 ft high and casts an 8 foot shadow, the oak tree casts a 12 foot shadow. How tall is the oak tree

Fudgin [204]

Answer: 4.8 ft

To answer this question you need to know how much shadow: actual height ratio. Flagpole is having 8 ft shadow with 20 ft actual height. The ratio should be= 8 ft: 20 ft= 0.4

Then multiply the ratio with the oak tree shadow. The equation would be:
0.4 x 12 ft= 4.8 ft

7 0

4 years ago

| 2(4+6) = how do I solve this?

podryga [215]

Answer:

Step-by-step explanation:

You use the distributive property.

Distribute the 2 to whatever is inside the parentheses. (You will multiply)

4×2=8

2×6=12

Substitute the answers above into the parentheses.

(8+12)=20