You start with the basic y=x^2.
That’s the parabola with a vertex at (0,0), opening up, etc.
The transformation aspect is the “+7” portion. This “+7” shifts the entire graph of y=x^2 up by 7 units.
The vertex is now (0,7), it still opens up, etc.
It’s kind of a silly question for your teacher to ask when the graph given only goes up to 6.
Answer:
for this problem you would have to find the coordients on the graph and then once you find all the coordients you would attach the lines together which should be a triangle.
Step-by-step explanation:
If you need answer's I can pin a picture
The the value of f(n)=
Step-by-step explanation:
The Arithemetic sequence is given as a,a+d,a+2d......a(n-1)d
and f(n)=a(n-1)d
The given sequence is 1/3,4/3,7/3...
By comapring with Arithemetic sequence,
we get a=1/3
a+d=4/3
1/3 + d = 4/3
d=1
To find value of f(10):
take n=10
f(n)=
f(n)=
f(n)=
f(n)=
Answer:
add x
Step-by-step explanation:
The operation "subtract 5" has eliminated the constant from the right side of the equation, so the next step would be to eliminate the variable from the left side. You do that by adding the opposite of the variable term.
The opposite of the term -x is x, so we "add x" as the next step.
This makes the solution look like ...
-x +6 = 5 -3x
-x +1 = -3x . . . . . . subtract 5
1 = -2x . . . . . . . . . add x
-1/2 = x . . . . . . . . .divide by -2
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<em>Comment on this solution</em>
My preference is always to add the opposite of the variable term with the smallest coefficient. Here, that term is -3x, so I would add 3x as a first step. This leaves the only x-term with a <em>positive</em> coefficient, so can reduce errors in the solution.
