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valentina_108 [34]
3 years ago
13

Help ASAP please I am desperate

Mathematics
1 answer:
snow_lady [41]3 years ago
7 0

Answer:

I think the first one is 18 and I think the second one is the 3rd one

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Order the numbers from least to greatest.
Mars2501 [29]

Answer:

4.446 then 4.464 then 5.228

Step-by-step explanation:

Hope this HELPS :D

7 0
3 years ago
Read 2 more answers
I do not know how the 1's cancel out to give 1/3^x
muminat

Unfortunately your teacher is using x as both a variable and a multiplication sign. This is something that can be avoided by using something like the asterisk symbol to indicate multiplication.

Anyways, notice how the expression 3^x*2^{2x}+1 shows up twice. Once in the numerator (the entire numerator) and once again in the denominator (nearly the whole thing)

Let's replace that messy expression with the variable y. So we're letting y = 3^x*2^x+1

This means,

\frac{3^x*2^2+1}{3^x\left(3^x*2^2+1\right)} = \frac{y}{3^x*y}

At this point you can probably see how to get \frac{1}{3^x} from here. The y terms cancel out when we divide leaving 1 up top and 3^x down below.

6 0
3 years ago
Ill give brainliest pls
Alona [7]

The answer is 207.

Search up the site Mathaway. It gives you answers.

6 0
3 years ago
PLEASE HELP :(((((((((
ziro4ka [17]
The first box would be 3.14 then second box would be the pi symbol and the third box would be 22/7
4 0
3 years ago
What is the smallest number which when decreased by 8 is divisible by 21, 27, 33, and 55?.
evablogger [386]

The required number which when decreased by 8 is divisible by 21, 27, 33, and 55 is 10403.

We need to find the smallest number which when decreased by 8 is divisible by 21, 27, 33, and 55.

<h3>What is LCM?</h3>

The least common multiple is also known as LCM (or) the lowest common multiple in math. The least common multiple of two or more numbers is the smallest number among all common multiples of the given numbers.

We find the LCM of 21, 27, 33, and 55.

3 | 21   27   33   55

3 | 7     9     11    55

3 | 7     3     11    55

7 | 7     1     11    55

5 | 1     1     11    55

11 | 1     1     11    11

 | 1     1     1     1

When decreased by 8, we have to add 8 in the LCM of the numbers.

That is 10395+8=10403

Therefore, the required number when decreased by 8 is divisible by 21, 27, 33, and 55 is 10403.

To learn more about the LCM of numbers visit:

brainly.com/question/10942748.

#SPJ4

8 0
2 years ago
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