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a_sh-v [17]
3 years ago
9

The values of x and y vary directly and one pair of values are given. Write an equation that relates x and y. x=1,y=0.5

Mathematics
1 answer:
professor190 [17]3 years ago
5 0

Answer:

y = 0.5x

Step-by-step explanation:

Given that x and y vary directly then the equation relating them is

y = kx ← k is the constant of variation

To find k use the condition x = 1, y = 0.5

k = \frac{y}{x} = \frac{0.5}{1} = 0.5

y = 0.5x ← equation of variation

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Step-by-step explanation:

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100:59

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The parabola below is a graph of the equation, . -y^2+x=-4
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The parabola divises the plan into 2 parts. Part 1 composes the point A, part 2 composes the points C, D, F.
+ All the points (x;y) satisfies: -y^2+x=-4 is on the <span>parabola.
</span>+ All the points (x;y) satisfies: -y^2+x< -4 is in part 1.
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3 years ago
Given the domain (-2,2,4) what is the range for the relation 3x-y=3
Bumek [7]

Answer:

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Step-by-step explanation:

First you have to put the relation in terms of y ⇒ 3x - y = 3⇒ 3x -3 = y

⇒ y = 3x - 3.

Then you replace the values indicated by the domain to find their "y" values (the ones that constitute the range).

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Evaluate the expression if a=2,b=-3,C=-1, and D=4<br><br><br> -2(b^2-5c)
Alenkasestr [34]

\qquad \qquad  \bf \huge\star \:  \:  \large{  \underline{Answer} }  \huge \:  \: \star

  • Equivalent value = -28

\textsf{  \underline{\underline{Steps to solve the problem} }:}

\qquad❖ \:  \sf \: - 2( {b}^{2}  - 5c)

( put the values )

\qquad❖ \:  \sf \: - 2 \{( - 3) {}^{2}  - 5( - 1) \}

\qquad❖ \:  \sf \: - 2(9 -  (- 5))

\qquad❖ \:  \sf \: - 2(9 + 5)

\qquad❖ \:  \sf \: - 2 \times 14

\qquad❖ \:  \sf \: - 28

\qquad \large \sf {Conclusion}  :

  • -2(b² - 5c) = -28

7 0
1 year ago
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