This is what a Stem and Leaf plot work. Do you know how to use it?
<u>Given</u><u> </u><u>:</u><u>-</u>
- To graph the line with slope 7 and y intercept -7.
<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>
<u>Solution</u><u> </u><u>:</u><u>-</u>
Here since the slope of the line is 7 and y intercept is -7 , we can use the slope intercept form of the line to find the equation of the line . The slope intercept form of the line is ,
y = mx + c
On putting the respective values ,
y = 7(x) + (-7)
y = 7x - 7
<u>For</u><u> </u><u>the</u><u> </u><u>graph</u><u> </u><u>see </u><u>attachment</u><u> </u><u>.</u>
Answer:
mean - 12.5
SD - 12.0
Step-by-step explanation:
Answer: hello, your answer is -0.45
Problem 1)
AC is only perpendicular to EF if angle ADE is 90 degrees
(angle ADE) + (angle DAE) + (angle AED) = 180
(angle ADE) + (44) + (48) = 180
(angle ADE) + 92 = 180
(angle ADE) + 92 - 92 = 180 - 92
angle ADE = 88
Since angle ADE is actually 88 degrees, we do NOT have a right angle so we do NOT have a right triangle
Triangle AED is acute (all 3 angles are less than 90 degrees)
So because angle ADE is NOT 90 degrees, this means
AC is NOT perpendicular to EF-------------------------------------------------------------
Problem 2)
a)
The center is (2,-3) The center is (h,k) and we can see that h = 2 and k = -3. It might help to write (x-2)^2+(y+3)^2 = 9 into (x-2)^2+(y-(-3))^2 = 3^3 then compare it to (x-h)^2 + (y-k)^2 = r^2
---------------------
b)
The radius is 3 and the diameter is 6From part a), we have (x-2)^2+(y-(-3))^2 = 3^3 matching (x-h)^2 + (y-k)^2 = r^2
where
h = 2
k = -3
r = 3
so, radius = r = 3
diameter = d = 2*r = 2*3 = 6
---------------------
c)
The graph is shown in the image attachment. It is a circle with center point C = (2,-3) and radius r = 3.
Some points on the circle are
A = (2, 0)
B = (5, -3)
D = (2, -6)
E = (-1, -3)
Note how the distance from the center C to some point on the circle, say point B, is 3 units. In other words segment BC = 3.