Answer:
<em>C. 94</em>
Step-by-step explanation:
Take in account for PEMDAS: Parenthesis, Exponent, Multiplication/Division, Addition/Subtraction.
First is to find the answer within the innermost parenthesis:
(4²+1).
Find the exponent of 4², which is 16, leaving you the equation of
16+1, which equals to 17.
Next you multiply 3 by 17 (since the parenthesis of 17 borders 3):
17×3=51.
Next, multiply 51 by 2 (same rule as 3×17).
51×2=102.
After this you get 102-2³. Now, find 2³=2×2×2=8
This gets you to 102-8, ending up with your answer of 94.
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
1:SPLIT THE MID FACTOR
<span>X^2+8X+12
</span><span> X^2+6X +2X+12 [12=6*2
</span><span> 8=6+2
</span>
2:TAKE COMMON FACTORS FROM 1ST TWO AND <span>LAST TWO FACTORS
</span><span>X^2+6X +2X+12 EQUALS
</span><span>X(X+6) + 2(X+6)
</span>
<span>3: TAKE (X+6) AS COMMON FACTOR
</span><span> X(X+6)+2(X+6) EQUALS
</span><span>(X+6) {X+2}</span>
Answer:
Step-by-step explanation:
your right on 105 cubic feet
Answer:
Possible, but very unlikely
Step-by-step explanation:
In a school of 150 students, the probability that at least two people have the same birthday is possible but very unlikely.
In a year, we have 365 total possible birthdays. If one student was born on 1st of January say, then the second student has 364 possible birth-dates assuming that they have different birthdays. This implies there is a higher probability that the second student has a different birthday.
Moreover, considering the school has only 150 students which is less than the total possible birth-days in a year, 365, the chances of two or more students sharing a birthday is possible but would be very unlikely.
Answer: tha answer is d
Step-by-step explanation: