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Gr equals 16 Br equals 8 and B is between G and R Is B the midpoint of segment Gr?
Wiki User
∙ 13y ago
Yes, because GB = GR - RB
Wiki User
∙ 13y ago
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What is the midpoint between 6 6 and 16 -6?
Midpoint = (6+16)/2 and (6-6)/2 = (11, 0)
What divided by what equal 16?
32 divided by 2 equals 16. 48divided by 3 equals 16. 64 divided by 4 equals 16. 16 divided by 1 equals 16.
What is 16-7 equals?
16-7 equals = 9
What is 16 percent equals in a fraction?
16% equals 4/25
16 plus 16 plus 16 equals equals?
that really easy how could anyone not know it equals 48 it islike walking
What is the class midpoint?
midpoint between 4-16
The class midpoint is?
midpoint between 4-16
What is the midpoint of the line segment with endpoints 16 5 and -6 -9?
If you mean endpoints of (16, 5) and (-6, -9) then its midpoint is (5, -2)
What is the midpoint between 6 6 and 16 -6?
Midpoint = (6+16)/2 and (6-6)/2 = (11, 0)
What is the midpoint of the class 16-19?
17.5
What is 16 shared between 2 equal?
16 shared between 2 equals 8
What is between the fraction 1 8 and 1 4?
To find the fractions that are between 1/8 and 1/4, change the fractions to a like denominator. 1/8 equals 2/16 1/4 equals 4/16 3/16 is between 1/8 and 1/4.
What divided by what equal 16?
32 divided by 2 equals 16. 48divided by 3 equals 16. 64 divided by 4 equals 16. 16 divided by 1 equals 16.
What is difference between Near structure and far structure?
We haven't used near and far addressing for well over a decade. It was common in older systems where memory was addressed by segment and offset. For instance, on a 32-bit system we might use 16-bits to address the segment and 16-bits to address the offset within that segment. If we were referring to an offset within the current segment then we'd use a 16-bit near pointer, but if we needed to refer to another segment then we'd use a 32-bit far pointer. Today we use a normalised pointers.
How segment register are used?
The segment register in the 80806/8088 microprocessor contains the base address (divided by 16) of a region of memory. Since the register is 16 bits in size, there are 65,536 possible segment base addresses, ranging from 00000H to FFFF0H, in increments of 00010H.After address translation at the instruction level, the generated 16 bit offset is added to the selected segment register times 16 to generate a physical address between 00000H and FFFFFH. (If the offset and base go past FFFFFH, they wrap around back to 00000H.) Since the offset is also 16 bits in size, and since the overlap is only 4 bits (times 16), then each 64 kb segment overlaps by 16 bytes.There are four segment registers; CS, DS, ES, and SS, standing for Code Segment, Data Segment, Extra Segment, and Stack Segment.CS is used for opcode fetches. DS is used for normal data. ES is used for certain string operations as the destination address. SS is used for stack and frame (BP) data.The segment registers can be implicitly selected by context, or they can be explicitly selected with a segment prefix opcode.
What is 16-7 equals?
16-7 equals = 9
How do you find the midpoint the slope the perpendicular slope and the equation for the perpendicular bisector of the line segment joining the points of 3 5 and 7 7?
Midpoint = (3+7)/2, (5+7)/2 = (5, 6) Slope of line segment = 7-5 divided by 7-3 = 2/4 = 1/2 Slope of the perpendicular = -2 Equation of the perpendicular bisector: y-y1 = m(x-x1) y-6 =-2(x-5) y = -2x+10+6 Equation of the perpendicular bisector is: y = -2x+16
