Answer: -0.5.
Step-by-step explanation:
The constant of variation k for the direct variation is given by:-
The given table:
x f( x )
0 0
2 -1
4 -2
7 -3.5
Then,
Hence, the constant of variation k for the direct variation is -0.5.
Answer:
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Your question does not say what were your options, therefore I will answer generically: in order to understand if a point (ordered pair) is contained in a line, you need to substitute the x-component of the pair in the equation of the line and see if the calculations give you the y-component of the pair.
Example:
Your line is <span> y = 4/3x + 1/3
Let's see if <span>(0, 0) and (2, 3) </span>belong to this line
y</span> = <span>4/3·0 + 1/3 = 1/3 </span>≠ 0
Therefore, the line does not contain (0, 0)
y = 4/3·2 + 1/3 = 9/3 = 3
Therefore, the line contains (2, 3)
A = (h * b) / 2
A = 200
h = 4b
200 = (b(4b) / 2
200 * 2 = b * 4b
400 = 4b^2
400 / 4 = b^2
100 = b^2
sqrt 100 = b
10 = b ..... base is 10 cm
h = 4b
h = 4(10)
h = 40 <=== height is 40 cm
a+b+c=0
[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc]
[a^2+b^2+c^2+2ab+2ac+2bc=0]
[a^2+b^2+c^2=-(2ab+2ac+2bc)]
[a^2+b^2+c^2=-2(ab+ac+bc)] (i)
also
[a=-b-c]
[a^2=-ab-ac] (ii)
[-c=a+b]
[-bc=ab+b^2] (iii)
adding (ii) and (iii) ,we have
[a^2-bc=b^2-ac] (iv)
devide (i) by (iv)
[(a^2+b^2+c^2)/(a^2-bc)=(-2(ab+bc+ca))/(b^2-ac)]
