Do the data in the table represent a direct variation or an inverse variation? write an equation to model the data in the table. x 1 2 5 10
y 40 20 8 4
a. Direct variation; y=40x
b. Direct variation; y=(1/40)x
c. Inverse variation; xy=40
d. Inverse variation; xy=1/40
I feel so stupid for not knowing the answer, please explain it to me.
Answers
y = 40,20, 8, 4
a) y = 40x
y = 40(1) = 40 ; y = 40(2) = 80 ;
y = 40(5) = 200 ; y = 40(10) = 400
b) y = (1/40)x ;
y = (1/40)1 = 1/40 ; y = (1/40)2 = 1/20 ;
y = (1/40)5 = 1/8 ; y =(1/40)10 = 1/4
c) xy = 40 ⇒ y = 40/x
y = 40/1 = 40 ; y = 40/2 = 20
y = 40/5 = 8 ; y = 40/10 = 4
Inverse variation. Choice C. xy = 40 is the correct answer.
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Which expressions are equivalent to the one below? Check all that apply 21x/7x
Can you please help me and quick
Answers
first subtract the fractions, since the denominators are the same.
2/5 - 1/5 = 1/5
Then subtract the whole numbers.
7- 6 = 1
So your answer is 1 1/5
Hope this helps! :)
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7 ⅖ − 6 ⅕ = 1 ⅕ .
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Answers
Answer:
3x^3+6x^+8x+24+47/x-2
Step-by-step explanation:
Answers
Answers
Answer:
l=95
Step-by-step explanation:
Perimeter is all the sides add up. So the equation if you know two of the sides, it should be: 338=(2*74)+(2*l)
Answer:
Step-by-step explanation:
the perimeter is 2w plus 2l
so 338 = 74(2)+2l
338 = 148+2l
190=2l
l=95
your length would be 95 m
Answers
Answer:
Vertex form of the function will be f(x) = (x - 1)² + 3.
Step-by-step explanation:
Vertex form of a quadratic function is given by f(x) = a(x - h)² + k
where (h, k) is the vertex of the given parabola.
Now we will convert the function in the vertex form.
f(x) = x² - 2x + 1 + 3
= (x - 1)² + 3
Therefore, the vertex form of the function will be f(x) = (x - 1)² + 3
and the vertex will be (1, 3).
a) function is positive on (-∞,-5)
b) fuction is negative on (-5,3)
c) function is positive on (-∞,1)
d) function is negative on (3,∞)
Answers
Answer:
d) function is negative on (3,∞)
Step-by-step explanation:
The even degree and negative leading coefficient tell you that the function is negative as x ⇒ ±∞. (Selections A and C cannot be correct.)
The odd multiplicity tells you the function crosses the x-axis at x=-5 and x=3, so will be non-negative between those values. (Selection B cannot be correct.)
The function is negative on (3, ∞).
Answer:
The graph of the function is positive on (-co, -5).
The graph of the function is negative on (3,co).
Step-by-step explanation:
We know that the roots are in: -5, 1 and 3.
and after 3, the graph is in the negative side, so between 1 and 3 the graph must be in the positive side, between -5 and 1 the graph must be in the negative side, and between -inifinity and -5 the graph must be in the positive side:
So the statements that are true are:
The graph of the function is positive on (-co, -5).
The graph of the function is negative on (3,co).