Answer:
x = 12
Step-by-step explanation:
The mistake made here is in making 130 = 9x + 7. They are not the same.
The central angle = 130 degrees.
Any angle sharing the endpoints of the central angle and having it's vertex on the same side of the central angle is 1/2 the central angle.
1/2 the central angle = 130 /2 = 65
Cyclic quadrilaterals (those with all 4 points touching a circle's circumference) have opposite angles supplementary.
65 + 9x + 7 = 180 Combine like terms on the left
72 + 9x = 180 Subtract 72 from both sides
72 - 72 + 9x = 180-72 Combine
9x = 108 Divide by 9
9x/9 = 108/9
x = 12
Answer:
Answer in the explanation section
Step-by-step explanation:
Even numbers create symmetry, but odd numbers create interest. An odd number of details is more effective at capturing your gaze. Odd numbers force your eyes to move around the grouping and by extension, the room. That forced movement is the heart of visual interest.
An even number is a number that can be divided into two equal groups. An odd number is a number that cannot be divided into two equal groups. Even numbers end in 2, 4, 6, 8 and 0 regardless of how many digits they have (we know the number 5,917,624 is even because it ends in a 4!). Odd numbers end in 1, 3, 5, 7, 9.
To me i think Odd numbers have more power
if this is wrong i'm sorry
Hope this helps
Answer:
the slope is stepper and the line is shifted up
Answer:
the lower right matrix is the third correct choice
Step-by-step explanation:
Your problem statement shows that you have correctly selected the matrices representing the initial problem setup (middle left) and the problem solution (middle right).
Of the remaining matrices, the upper left is an incorrect setup, and the lower left is an incorrect solution matrix.
__
We notice that in the remaining matrices on the right that the (2,3) term is 0, and the (3,2) and (3,3) terms are both 1.
The easiest way to get a 0 in the 3rd column of row 2 is to add the first row to the second. When you do that, you get ...
![\left[\begin{array}{ccc|c}1&1&1&29000\\1+2&1-3&1-1&1000(29+1)\\0&0.15&0.15&2100\end{array}\right] =\left[\begin{array}{ccc|c}1&1&1&29000\\3&-2&0&30000\\0&0.15&0.15&2100\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%261%2629000%5C%5C1%2B2%261-3%261-1%261000%2829%2B1%29%5C%5C0%260.15%260.15%262100%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%261%2629000%5C%5C3%26-2%260%2630000%5C%5C0%260.15%260.15%262100%5Cend%7Barray%7D%5Cright%5D)
Already, we see that the second row matches that in the lower right matrix.
The easiest way to get 1's in the last row is to divide that row by 0.15. When we do that, the (3,4) entry becomes 2100/0.15 = 14000, matching exactly the lower right matrix.
The correct choices here are the two you have selected, and <em>the lower right matrix</em>.
Answer:
75806
Step-by-step explanation:
82,277-6,471=75806
