Answer:
1. Two segments are MC and AX
2. Point N
3. Three collinear points are points A, point N and point X
4. Two other names for V are quadrilateral and rectangle
5. No N, M, and X are in a different plane from R
6. Two rays are ray NR and ray NA
7. The pair of opposite rays with end point N are ray NA and ray NX and ray NM and ray NC
8. Three lines are shown in the figure
Step-by-step explanation:
1) A segment is a line portion with two end points
2) A point marks a position
3) Collinear points are points on the same line
3) A quadrilateral is a four sided figure while rectangle has 90° interior angles
4) A ray is an line extending from a point without ending
5) Opposite rays are rays extending from the same point but in opposite directions
6) A line is a straight one dimensional geometric figure that has no width but infinite number of points extending without end in both directions
Answer:
Equation I , II and V.
Step-by-step explanation:
Number II:L
0.5(8x + 4) = 4x + 2 Distributing the 0.5 over the parentheses:
4x + 2 = 4x + 2
The 2 sides of this equation are identical so we can make x any value to fit this identity. That is there are Infinite Solutions.
I and V are also included:
I: 12x + 24 = 12x + 24.
V left side = right side.
Answer:
2 2/5
Step-by-step explanation:
12 can go into 5 2 times with a left over of two, so it would be 2 which is 10 and a left over of two. The denominator stays the same. So it would be 2 2/5.
Answer:
The answer is 140
Step-by-step explanation:
When the total measurements of the six-shaped corners is equal 720
Then x equal to

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The only way 3 digits can have product 24 is
1 x 3 x 8 = 241 x 4 x 6 = 242 x 2 x 6 = 242 x 3 x 4 = 24
So the digits comprises of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4
To be divisible by 3 the sum of the digits must be divisible by 3.
1+ 3+ 8=12, 1+ 4+ 6= 11, 2 +2 + 6=10, 2 +3 + 4=9Of those sums of digits, only 12 and 9 are divisible by 3.
So we have ruled out all but integers whose digits consist of1,3,8, and 2,3,4.
Meanwhile they must be odd they either must end in 1 or 3.
The only ones which can end in 1 are 381 and 831.
The others must end in 3.
They must be greater than 152 which is 225. So the
First digit cannot be 1. So the only way its digits can contain of1,3,8 and close in 3 is to be 813.
The rest must contain of the digits 2,3,4, and the only way they can end in 3 is to be 243 or 423.
So there are precisely five such three-digit integers: 381, 831, 813, 243, and 423.