Answer:
On Monday the temperature was 35 - 4 = 31°. On Tuesday, it was 31 + 2 = 33° and on Wednesday it was 33 - 5 = 28° F.
The correct answer is D.14.
The interquartile range is the difference between the Upper Quartile and the Lower Quartile of a data set.
By arranging the numbers in the stem and leaf plot from lowest to highest value, you get an even set of data (31,33,35,41,43,46,48,49,49,50). By dividing the set into two, you get a lower half of your data (31,33,35,41,43) and an upper half (46,48,49,49,50). The lower quartile is the median or the midpoint of the lower half of your data while upper quartile is for the upper half. The lower quartiles is 35 and upper quartile is 49. The difference between the two is 14.
Answer:
All the sides of the triangle X'Y'Z' are twice as long as the sides of the original triangle XYZ. The triangle XYZ has been enlarged by a scale factor of 2.
All the sides of the triangle X'Y'Z' are twice as long as the sides of the original triangle XYZ. The triangle XYZ has been enlarged by a scale factor of 2.Enlargement is an example of a transformation. A transformation is a way of changing the size or position of a shape.
All the sides of the triangle X'Y'Z' are twice as long as the sides of the original triangle XYZ. The triangle XYZ has been enlarged by a scale factor of 2.Enlargement is an example of a transformation. A transformation is a way of changing the size or position of a shape.To enlarge a shape, a centre of enlargement is required. When a shape is enlarged from a centre of enlargement, the distances from the centre to each point are multiplied by the scale factor.
All the sides of the triangle X'Y'Z' are twice as long as the sides of the original triangle XYZ. The triangle XYZ has been enlarged by a scale factor of 2.Enlargement is an example of a transformation. A transformation is a way of changing the size or position of a shape.To enlarge a shape, a centre of enlargement is required. When a shape is enlarged from a centre of enlargement, the distances from the centre to each point are multiplied by the scale factor.The lengths in triangle A'B'C' are three times as long as triangle ABC. The distance from O to triangle A'B'C' is three times the distance from O to ABC.
Answer:
no it is not perpedicular
Step-by-step explanation:
I plugged the equation into a graphing calculator and the line does not appear to be perpendicular
Answer:
1 x=-2.5 y = -5.5
2. x=5 y=1
Step-by-step explanation:
1) What is the solution of the given system?
5x-y=-7
3x-y=-2
Multiply the second equation by -1
-1*(3x-y)=-1(-2)
-3x +y = 2
Now add the first equation to the modified second equation
5x-y=-7
-3x +y = 2
------------------
2x = -5
Divide each side by 2
2x/2 = -5/2
x = -2.5
Now we need to find y
-3x+y =2
-3(-2.5) +y =2
7.5 +y =2
Subtract 7.5 from each side
7.5 -7.5 +y =2-7.5
y = -5.5
2) what is the solution of the given system?
5x+7y=32
8x+6y=46
Divide the second equation by 2
8x/2+6y/2=46/2
4x+3y =23
Multiply the first equation by 4
4 (5x+7y)=32*4
20x+28y = 128
Now multiply the modified 2nd equation by -5
-5(4x+3y )=-5(23
)
-20x -15y = -115
Lets add the new equations together to eliminate x
20x+28y = 128
-20x -15y = -115
---------------------
13y = 13
Divide each side by 13
13y/13 =13/13
y=1
Now substitute back in to find x
5x+7y=32
5x +7(1) =32
5x +7 =32
Subtract 7 from each side
5x+7-7 =32-7
5x =25
Divide by 5
5x/5 =25/5
x=5
