You are correct, 153 in^2.
Yo sup??
Let the total marks obtained by the 16 students be S
77=S/16
S=1232
When Taqeesha takes the test let her score be x
78=S+x/17
1326=1232+x
x=94
Therefore the answer is option E.94
Hope this helps.
Parent Function: f(x)=x^2
Horizontal Shift: Right 5 Units
Vertical Shift: Up 3 Units
Reflection about the x-axis: None
Reflection about the y-axis: None
Vertical Compression or Stretch: None
Answer:
The equation of perpendicular bisector of the line segment passing through (-5,3) and (3,7) is: ![y = -2x+3](https://tex.z-dn.net/?f=y%20%3D%20-2x%2B3)
Step-by-step explanation:
Given points are:
(-5,3) and (3,7)
The perpendicular bisector of line segment formed by given points will pass through the mid-point of the line segment.
First of all we have to find the slope and mid-point of given line
Here
(x1,y1) = (-5,3)
(x2,y2) = (3,7)
The slope will be:
![m = \frac{y_2-y_1}{x_2-x_1}\\m = \frac{7-3}{3+5}\\m = \frac{4}{8}\\m = \frac{1}{2}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7By_2-y_1%7D%7Bx_2-x_1%7D%5C%5Cm%20%3D%20%5Cfrac%7B7-3%7D%7B3%2B5%7D%5C%5Cm%20%3D%20%5Cfrac%7B4%7D%7B8%7D%5C%5Cm%20%3D%20%5Cfrac%7B1%7D%7B2%7D)
The mid-point will be:
![(x,y) = (\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2})\\ = (\frac{-5+3}{2},\frac{3+7}{2})\\= (\frac{-2}{2},\frac{10}{2})\\=(-1,5)](https://tex.z-dn.net/?f=%28x%2Cy%29%20%3D%20%28%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D%20%2C%20%5Cfrac%7By_1%2By_2%7D%7B2%7D%29%5C%5C%20%3D%20%28%5Cfrac%7B-5%2B3%7D%7B2%7D%2C%5Cfrac%7B3%2B7%7D%7B2%7D%29%5C%5C%3D%20%28%5Cfrac%7B-2%7D%7B2%7D%2C%5Cfrac%7B10%7D%7B2%7D%29%5C%5C%3D%28-1%2C5%29)
Let m1 be the slope of the perpendicular bisector
Then using, "Product of slopes of perpendicular lines is -1"
![m.m_1 = -1\\\frac{1}{2}.m_1 = -1\\m_1 = -1*2\\m_1 = -2](https://tex.z-dn.net/?f=m.m_1%20%3D%20-1%5C%5C%5Cfrac%7B1%7D%7B2%7D.m_1%20%3D%20-1%5C%5Cm_1%20%3D%20-1%2A2%5C%5Cm_1%20%3D%20-2)
We have to find the equation of a line with slope -2 and passing through (-1,5)
The slope-intercept form is given by:
![y = mx+b\\y = -2x+b](https://tex.z-dn.net/?f=y%20%3D%20mx%2Bb%5C%5Cy%20%3D%20-2x%2Bb)
Putting the point (-1,5) in the equation
![5 = -2(-1)+b\\5 = 2+b\\b = 5-2\\b = 3](https://tex.z-dn.net/?f=5%20%3D%20-2%28-1%29%2Bb%5C%5C5%20%3D%202%2Bb%5C%5Cb%20%3D%205-2%5C%5Cb%20%3D%203)
The final equation is:
![y = -2x+3](https://tex.z-dn.net/?f=y%20%3D%20-2x%2B3)
Hence,
The equation of perpendicular bisector of the line segment passing through (-5,3) and (3,7) is: ![y = -2x+3](https://tex.z-dn.net/?f=y%20%3D%20-2x%2B3)
Answer: The numbers on a distance of 3.2 units from −4 on a number line are -7.2 and - 0.8 .
Step-by-step explanation:
The numbers on a number line are from - infinity to + infinity whose center is at 0.Om left sides of 0 , there are negative integers and on the right side there are positive integers.
So, The numbers on a distance of 3.2 units from −4 on a number line are
-4 - 3.2 and -4+3.2
= - (4+3.2) and - (4-3.2)
= -7.2 and - 0.8
Hence, the numbers on a distance of 3.2 units from −4 on a number line are -7.2 and - 0.8 .