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Murrr4er [49]
3 years ago
12

Suppose y varies directly with x. if y=-3 when x=5 find y when x=-1

Mathematics
2 answers:
nadezda [96]3 years ago
4 0

Answer:

The value of y at x=-1 is 3/5.

Step-by-step explanation:

It is given that y varies directly with x. It means y is directly proportional to x.

y\propto x

y=kx              .... (1)

where, k is constant of proportionality.

It is given that y=-3 when x=5.

Substitute y=-3 and x=5 in equation (1).

-3=k(5)

-\frac{3}{5}=k

The value of k is -3/5.

y=-\frac{3}{5}x          .... (2)

We need to find the value of y when x=-1.

Substitute x=-1 in equation (2).

y=-\frac{3}{5}(-1)

y=\frac{3}{5}

Therefore the value of y at x=-1 is 3/5.

Scrat [10]3 years ago
3 0
\bf \begin{array}{cccccclllll}
\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\
\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\
y&=&{{ k}}&\cdot&x

&&  y={{ k }}x
\end{array}
\\ \quad \\
\textit{we know that, when}
\begin{cases}
y=-3\\
x=5
\end{cases}\implies y=kx\implies (-3)=k(5)

solve for "k", to find the "constant of variation",
then plug it back in the y = kx, to get the equation
now
what's is y when x = -1?
well, just plug that in the equation with the found "k" value,
to get "y" :)
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By using the concept of uniform rectilinear motion, the distance surplus of the average race car is equal to 3 / 4 miles. (Right choice: A)

<h3>How many more distance does the average race car travels than the average consumer car?</h3>

In accordance with the statement, both the average consumer car and the average race car travel at constant speed (v), in miles per hour. The distance traveled by the vehicle (s), in miles, is equal to the product of the speed and time (t), in hours. The distance surplus (s'), in miles, done by the average race car is determined by the following expression:

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