Numbers 1 and 2 are correct.
Question 3:
We can see in each figure, the number of rows matches the figure amount. For example, figure 1 has 1 row, figure 2 has 2 rows, figure 3 has 3 rows, etc. Another thing we can see is that one block gets added to each column in every figure. If we count starting from 3 and keep adding 1 for each column, we get 22 blocks for one column. In figure 20, there will be 20 rows and 22 columns. We also have to keep in mind that they add 2 extra blocks at the top and bottom that stick out like the ones shown. In total, figure 20 will consist of 442 blocks.
Question 4:
[(2 + f) * (f)] + 2 = b
(2 + f) represents the number of blocks in each column.
(f) represents the number of blocks in each row.
2 represents the two extra blocks in each figure.
Note that f in this equation equals the number of the figure. For example, figure 3 is f = 3
We can test our equation with figure three.
[(2 + f) * (f)] + 2 = b
We have figure three so f = 3. Substitute and solve.
[(2 + 3) * (3)] + 2 = b
[5 * 3] + 2 = b
15 + 2 = b
17 = b
In the example, the number of blocks is 17. Therefore, this equation works for any figure.
Best of Luck!
The total cereal mixture is 45 kilograms
<em><u>Solution:</u></em>
Given that, cereal mixture contains Almond, Wheat, and Corn in the ratio of 3:5:7
Therefore,
Almond : Wheat : Corn = 3 : 5 : 7
Let the almond present be 3x
Let the wheat present be 5x
Let the corn present be 7x
Therefore,
Total cereal mixture = 3x + 5x + 7x
Total cereal mixture = 15x
<em><u>Given that mixture contains 9 kg of almond</u></em>
Almond = 9
We know that,
Almond present = 3x
3x = 9
x = 3
Therefore,
Total cereal mixture = 15x = 15(3) = 45
Thus total cereal mixture is 45 kilograms
Answer:
12 and 9
Step-by-step explanation:
x=larger number y=smaller number
x+y=21
x=2y-6 (system of equations)
x=21-y
2y-6=21-y
3y=27
y=9
x+9=21
x=12
X=2 I did the test and I got it correct
Answer:
mACE = 1/2 mACD
because Ray CE is the angle bisector of ACD