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Scott, MIT Graduate
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1.write equation of the line containing the given point and
Resolved Question:
1.write equation of the line containing the given point and parallel to the given line
Express your answer in the form y = mx + b (3,6); x +9y =7
2. the function W(d)0.112d approximates the amount in centimeter of water that results from d cm of snow melting. Find the amount of water the results from snow melting from depths of 20cm,38cm,and 85cm.
3 Given the K(t)= 6t +1 find each of the following
K(4)
K(6)
K(6.3)
4 Determine whether the graphs of the pair of lines are parallel
X+9=y
yx=5
what is the slope of the line x+9=y
what is the slope of the lineyx=5
are the graphs of given equation parallel?
5 Use the graph of the function f to find
F(0) F(1) F(2)
6 Find the slope of the line (0,5) (5,7)
The slope of the line is m=
7 Graph the equation by plotting points
Y=2
8 Find the slope if it exits, of the line containing the pair of points
(8,9) and (12,19)
The slope m=
9 Find an equation of the line having the given slope and containing the given point.
M=3,(1,6)
10 Find the domain
P(x)= Xto the second power + 8
11 For the graph, determine the value of
F(1)
The domain is
F(x)= 4
Y=
The range is
12 Find the slope and the y intercept
F(x) = 10x6
13 Find the indicated out puts for f(x) = 4x to second power 5x
F(0)
F(1)
F(2)
14 find an equation of the line containing the given pair of points. Express your answer in the form x=a,y=b,or y= mx +b
(1,1) and (4,4)
What is an equation of the line ?
Y=
15 determine whether the graph of the two equations are perpendicular.
Y= 6x4
9y = 5x
16 are the graphs of the given equations perpendicular graph using slope and the yintercept
X +2y = 6
Find slope and the y intercept
61/2y = 4x
Submitted:
6 years ago.
Category:
Homework
Expert:
Scott
replied 6 years ago.
Hi there!
Do you have graphs for #5 and #11?
I can do all the other ones without graphs, but I'll need them to do those two.
Thanks,
Scott
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Customer:
replied 6 years ago.
dont worry about those two then
Expert:
Scott
replied 6 years ago.
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