What's the answer ? I'm almost positive you do multiplication first then add but not I'm not sure . 12+34+5*6= 44

Answers

Answer 1

Answer: PEMDAS
multiply first

1*2=2
3*4=12
5*6=30

now we have
2+12+30=44
add
2+12=14
14+30=44
44=44
true

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Identify the 17th term of a geometric sequence where a1 = 16 and a5 = 150.06. Round the common ratio and 17th term to the nearest hundredth.

Answers

The geometric sequence formula is expressed as an = a1 * r^(n-1) where n is an integer. In this case, upon substitution, 150.06 = 16 * r^(4). extracting r, r is equal to 1.75. Hence the 17th term from the formula is equal to 123802.32.

Answer:

a17 ≈ 123,802.31

Step-by-step explanation:

Gender: Grade 9 Grade 10 Grade 11 Grade 12Male: 53 19 27 37

Female: 31 42 50 61

What percentage of females are in grades 11 and 12? Round to the nearest percent.

27%

35%

60%

67%

Answers

Answer:

Option C- 60%

Step-by-step explanation:

Given :

Gender : Grade 9 Grade 10 Grade 11 Grade 12

Male: 53 19 27 37

Female: 31 42 50 61

To find : Percentage of females are in grades 11 and 12

Solution: Total number of females= 31+42+50+61= 184

Number of females in grades 11 and 12 = 50+61 = 111

Percentage of females are in grade 11 and 12 = (No. of females in 11 and 12/total no. of females) × 100

$\Rightarrow (111)/(184)*100=(11100)/(184)=60.32$

Percentage is approx. = 60%

Therefore, option C is correct

There are 31+42+50+61=184 total females, and 50+61=111 females in grades 11 and 12, so the percentage is 111/184*100=60.33 percent, which can be rounded to 60%.

What is x4-81 factored?

Answers

What we need here:

"The difference of two squares is the same as (their sum) times (their difference)."

x^4 is the square of x² .
81 is the square of 9 .
Their difference is the same as (x² + 9) times (x² - 9).

But (x² - 9) is also the difference of two squares,
so it's the same thing as (x+ 3) times (x-3).

So the complete factored form of the original expression is

(x^4 - 81) = (x² + 9) (x+3) (x - 3)

$x^4-81=(x^2)^2-9^2=(x^2-9)(x^2+9)=(x^2-3^2)(x^2+9)=\n\n=(x-3)(x+3)(x^2+9)$

A password contains exactly 5 letters how many passwords are possible if letters cannot be used more than once?A. 3120

B. 7,893,600

C. 65,780

D. 11,881,376

Answers

The first letter can be any one of the 26 in the alphabet.
   For each one ...
The 2nd letter can be any one of the remaining 25.
   For each one ...
The 3rd letter can be any one of the remaining 24.
   For each one ...
The 4th letter can be any one of the remaining 23.
   For each one ...
The 5th letter can be any one of the remaining 22.

The total number of possibilities is

   (26 · 25 · 24 · 23 · 22) = 7,893,600 .

Hello please help thanks