Respuesta :
The slope-intercept form of a line is:
y=mx+b, where m=slope and b=y-intercept.
First find the slope using the two points clearly marked.
Slope or velocity is (y2-y1)/(x2-x1), in this case:
m=(-2-0)/(2--6)=-2/8=-1/4, so far we now have:
y=-0.25x+b, now using either point we can solve for b, using (2,-2):
-2=-0.25(2)+b
-2=-0.5+b
-1.5=b, so the line is:
y=-0.25x-1.5
So the slope is -1/4 and the y-intercept is -1.5
....
For the field of dimensions that are 3.2 times that of the playground...
perimeter=2(L+W) for the playground.
If the dimensions of the field are 3.2 times greater then:
p=2(3.2L+3.2W)
p=(3.2)(2(L+W))
So the perimeter of the field is 3.2 times that of the playground, so statements c and d are not true....
Now for the areas...the area of the playground is:
A=xy and the area of the field is then:
A=(3.2x)(3.2y)
A=10.24xy
So the area of the field is 10.24 times that of the area of the playground.
Thus B is the only true statement.
y=mx+b, where m=slope and b=y-intercept.
First find the slope using the two points clearly marked.
Slope or velocity is (y2-y1)/(x2-x1), in this case:
m=(-2-0)/(2--6)=-2/8=-1/4, so far we now have:
y=-0.25x+b, now using either point we can solve for b, using (2,-2):
-2=-0.25(2)+b
-2=-0.5+b
-1.5=b, so the line is:
y=-0.25x-1.5
So the slope is -1/4 and the y-intercept is -1.5
....
For the field of dimensions that are 3.2 times that of the playground...
perimeter=2(L+W) for the playground.
If the dimensions of the field are 3.2 times greater then:
p=2(3.2L+3.2W)
p=(3.2)(2(L+W))
So the perimeter of the field is 3.2 times that of the playground, so statements c and d are not true....
Now for the areas...the area of the playground is:
A=xy and the area of the field is then:
A=(3.2x)(3.2y)
A=10.24xy
So the area of the field is 10.24 times that of the area of the playground.
Thus B is the only true statement.