Answer:
m
Step-by-step explanation:
Answer:
The area is 21 inches
Step-by-step explanation:
3x3=9
3x2=6
9+6=15 ( inches in the white box )
6x6=36 ( area of the whole box)
36-15= 21 ( inches in the shaded area )
Answer:
{y≥1
,{y-x>0
Step-by-step explanation:
First of all you have to consider the shaded region. It is bound by two lines.
The first line is a solid line that cuts the y-axis at +1. it's equation is y = 1. since the shade region is on the upper side where y values increase, the unequivocally will be y≥1. notice that the sign ≥ is due to the solid line which indicates points on the solid line are part of the solution.
the second line is the broken line. it passes through the origin (0,0) and (1,1) any two points can be taken. the gradient is 1. m= (y1-y2)/(x1-x2) = (0-1)/(0-1)=(-1/-1)= 1. the equation of a straight line is
y=mx + c where m is gradient and c is the VA)ue of y as the line crosses the y axis ( y-intercept) which in this case is 0 at (0,0).so the equation will be y=1(x) + 0
y=x if we subtract x from both sides we have
y-x=0
since the shaded region is on the upper side as y-x increases the in equality will be
y-x>0 notice since the line is broken it shall be just > not≥ because points on a broken line are not included in the shaded region.
Given that Sienna planned a rectangular pool drawn on a scale 11 inches long by 7 inches wide, and Sienna used the scale drawing above to create a pool that is 14 ft wide x 22 ft long, and she then decided to make a pool with a final length of 33 ft, to determine which expression finds the change in scale factor for the longer pool Sierra is building the following calculation must be performed:
- 11 inches = 0.916 feet
- 11 / 0.916 = 12
- 7 inches = 0.583333 feet
- 7 / 0.5833333 = 12
- Thus, the initial scale is 1:12.
- 33 / 0.916 = 36
Therefore, the scale factor went from 1:12 to 1:36.
Learn more in brainly.com/question/2302757
1. A straight line segment can be drawn joining any two points.
2. Any straight line segment can be extended indefinitely in a straight line.
3. Given any straight line segment, a circle can be drawn having the segment as radius and one endpoint as center.
4. All right angles are congruent.
5. If two lines are drawn which intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough. This postulate is equivalent to what is known as the parallel postulate.
