不是我要的 营养分, 一、加; 的 第一 年11 在公司 16 115 更多 11 ( - 日) 上海

PLEASE HELP {It costs 5,700 every one sixth of an hour to build a hotel how many cents does building a hotel cost per second?}

Keith_Richards [23]

One sixth of an hour is 10 minutes.
That means it costs 5,700 every 10 minutes.

Costs 5,700 = 10 minutes

Dive 10 on both sides.

Costs 570 = 1 minute

Lets convert the minute into seconds.

Costs 570 = 60 seconds

Divide both sides by 60.

Costs 9.5 cents = 1 second

Round it off since you can't pay 9.5 cents.

Costs 10 cents = 1 second

This means that every second they pay 10 cents for the hotel. :D

8 0

3 years ago

Helppppp!!!!! Plzzz!!!! ASAP!!!!!!! 20+PTS and brainliest!!!!!

garik1379 [7]

Answer:

Step-by-step explanation:

cos(46°) = 24/x

x = 24/cos(46°)

7 0

3 years ago

If bacteria took 20 minutes to double from 2 to 4, how many bacteria would there be after 1 hour

iren2701 [21]

4 * 2 * 2 * 2 = 32 bacteria

Hope it helps!

3 0

3 years ago

Read 2 more answers

Question 1,2, and 3 how do i factor those? Can you show the work and explain how?

DiKsa [7]

1: $3n^{2}+9n+6$

notice that each part is divisible by 3

$3n^{2}$ ÷ 3 = $n^{2}$

9n ÷ 3 = 3n

6 ÷ 3 = 2

so it becomes $3(n^{2} +3n+2)$

3n can be rewritten as 2n+n

-you want to rewrite it into two numbers that multiply to the number that's alone (in this case 2)

which would get you

$3(n^{2} +2n+n+2)$

Now that it's rewritten, you can factor out n + 2 from the equation.

the answer is

3(n+2)(n+1)

And you can check that by multiplying (n+2)(n+1) which is $n^{2} +2n+n+2$ and then each of those by 3, which is $3n^{2} +6n+3n+6$ or $3n^{2}+9n+6$ , our origional equation

2: $28+x^{2} -11x$

So I rewrote this as $x^{2} -11x+28$ (it's the same thing, just reordered using the commutative property)

now -11x can be rewritten as -4x-7x

(remember, the two numbers should multiply to equal 28, which is our constant.)

$x^{2} -4x-7x+28$

now we can factor out x from the first expression and -7 from the second

$x(x-4)-7(x-4)$

and lastly you factor out x-4,

which would give you

(x-4)(x-7)

Make sure to check your work and make sure it multiplies to $x^{2} -11x+28$

3: $9x^{2} -12x+4$

The first thing I notice when looking at this problem is that both 9 and 4 are perfect squares. Not only that, but they are the squares of 2 and 3, which are products of -12

So if you rewrite 9 as $3^{2}$ and 4 as $2^{2}$ , the equation becomes