Answer:
The fraction is 7/10 and the decimal is 0.7
Step-by-step explanation:
Each incriment represents 1/10, or 0.1.
Answer:
x²(9x– 11)(9x + 11)
Step-by-step explanation:
81x⁴ – 121x²
The expression can be factorised as follow:
81x⁴ – 121x²
x² is common to both term. Thus:
81x⁴ – 121x² = x²(81x² – 121)
Recall:
81 = 9²
121 = 11²
Therefore,
x²(81x² – 121) = x²(9²x² – 11²)
= x²[(9x)² – 11²]
Difference of two squares
x²(9x– 11)(9x + 11)
Therefore,
81x⁴ – 121x² = x²(9x– 11)(9x + 11)
Answer:
Cross section is the term name given to the front of any prism. SO, they are only found in a prism. They are recognized as the side/face that appears twice in a prism, and the shape have a visible length between these 2, known as the height
Step-by-step explanation:
Hope this helps
Answer:
Option B
Step-by-step explanation:
Looking at the options, option B is correct because when multiplying it by matrix A, it yields the matrix AB as follows;
First row of A multiplied by first column of matrix in option D;
(1 × -1) + (0 × 0) + (0 × 0) = -1 which corresponds to the first number on the first row of Matrix AB
Since majority of matrix AB are zero, I will just prove the ones that are not zero.
Thus;
Second row of matrix A is multiplied by second column of matrix in option D;
(0 × 0) + (-1 × -1) + (0 × 0) = 1 which is same as 2nd number on second row in matrix AB
Lastly, third row of matrix A is multiplied by third column of matrix in option D;
(0 × 0) + (0 × 0) + (1 × -1) = -1 which is same as third number in third row in matrix AB
(a) From the histogram, you can see that there are 2 students with scores between 50 and 60; 3 between 60 and 70; 7 between 70 and 80; 9 between 80 and 90; and 1 between 90 and 100. So there are a total of 2 + 3 + 7 + 9 + 1 = 22 students.
(b) This is entirely up to whoever constructed the histogram to begin with... It's ambiguous as to which of the groups contains students with a score of exactly 60 - are they placed in the 50-60 group, or in the 60-70 group?
On the other hand, if a student gets a score of 100, then they would certainly be put in the 90-100 group. So for the sake of consistency, you should probably assume that the groups are assigned as follows:
50 ≤ score ≤ 60 ==> 50-60
60 < score ≤ 70 ==> 60-70
70 < score ≤ 80 ==> 70-80
80 < score ≤ 90 ==> 80-90
90 < score ≤ 100 ==> 90-100
Then a student who scored a 60 should be added to the 50-60 category.
