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3 years ago

8

Which number is three hundred thousand times larger than 1.6 x 102 ? A) 4.8 x 106 B) 4.8 x 107 C) 4.8 x 108 D) 4.8 x 109

2 answers:

Bas_tet [7]3 years ago

8 0

Answer: D

I hope this helps and have a wonderful day filled with joy!!

-Timarra

Lapatulllka [165]3 years ago

7 0

I think the answer for you is d

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What is the rounding or compatible number<br><br> of 256+321

Nana76 [90]

<span>Compatible </span>numbers for 256...
1. Rounded to the nearest hundred = 200
2. Rounded to the nearest tens place = 260

Compatible numbers for 321...
1. Rounded to the nearest hundred = 300
2. Rounded to the nearest tens place = 320

I hope this helps! :)

4 0

3 years ago

Read 2 more answers

Use f(x) = x +5 and g(x) = 2x.

NemiM [27]

2x+10 is the correct answer.

4 0

3 years ago

Please help me with this one :(

dimulka [17.4K]

Answer:

I can assure you, that this is not that hard. but. I will help you with one of them.

Step-by-step explanation:

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it's basically a crowd of people.

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2 years ago

A bank loaned out $12,00, part of it at the rate of 9% per year and the rest at 19% per year. If the interest received in one

Crazy boy [7]

Answer:

$7,800 was invested at 9% and $4,200 at 19%.

Step-by-step explanation:

Given that a bank loaned out $ 12,000, part of it at the rate of 9% per year and the rest at 19% per year, if the interest received in one year totaled $ 1,500, to determine how much was loaned at 9% the following calculation must be performed:

12,000 x 0.09 + 0 x 0.19 = 1,080

10,000 x 0.09 + 2,000 x 0.19 = 1,280

8,000 x 0.09 + 4,000 x 0.19 = 1,480

7,900 x 0.09 + 4,100 x 0.19 = 1,490

7,800 x 0.09 + 4,200 x 0.19 = 1,500

Thus, $ 7,800 was invested at 9% and $ 4,200 at 19%.

6 0

3 years ago

What are the next 3 numbers in this pattern of integers -2, -1, -1, 0, -1, 1, -2

sashaice [31]

Answer:

The next 3 numbers to the given pattern of integers is -2,-1,-1,0,-1,1,-2 are 3,-5,8

Step-by-step explanation:

Given pattern of integers is -2,-1,-1,0,-1,1,-2

To find their next 3 numbers of the given pattern :

Let $a_{1}=-2$

$a_{2}=-1$

$a_{3}=a_{1}-a_{2}$

$=-2-(-1)$

$=-2+1$

Therefore $a_{3}=-1$

$a_{4}=a_{2}-a_{3}$

$=-1-(-1)$

$=-1+1$

Therefore $a_{4}=0$

Similarly we have $a_{5}=-1$ , $a_{6}=1$ , $a_{7}=-2$

The general form to the given pattern is

$a_{n}=a_{n-2}+a_{n-1}$ for $n={\{1,2,3,4,5,6,7,8,9,10}\}$

Therefore to find the $a_{8},a_{9},a_{10}$

Put n=8 in $a_{n}=a_{n-2}-a_{n-1}$

$a_{8}=a_{8-2}-a_{8-1}$

$=a_{6}-a_{7}$

$=1-(-2)$

$a_{8}=3$

Put n=9 in $a_{n}=a_{n-2}-a_{n-1}$

$a_{9}=a_{9-2}-a_{9-1}$

$=a_{7}-a_{8}$

$=-2-(3)$

$a_{9}=-5$

Put n=10 in $a_{n}=a_{n-2}-a_{n-1}$

$a_{10}=a_{10-2}-a_{10-1}$

$=a_{8}-a_{9}$

$=3-(-5)$

$a_{10}=8$

Therefore the next 3 numbers are 3,-5,8

4 0

3 years ago