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krok68 [10]
3 years ago
12

How to write as equation, the product of 7 and a number

Mathematics
2 answers:
mote1985 [20]3 years ago
8 0
Let us say that "a number" is x. If you were to do this, then the product of 7 and a number, x would be 7x.

BlackZzzverrR [31]3 years ago
5 0
You would consider it as "7n=x" as n is the undefined number and since it is an equation, it must have an equal sign, so x is the product.
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Answer:

The rate of change of the distance \frac{ds}{dt} when x = 9 and y = 12 is -7.6\:\frac{m}{s}.

Step-by-step explanation:

This is an example of a related rate problem. A related rate problem is a problem in which we know one of the rates of change at a given instant \frac{dx}{dt} and we want to find the other rate \frac{dy}{dt} at that instant.

We know the rate of change of x-coordinate and y-coordinate:

\frac{dx}{dt}=-2\:\frac{m}{s} \\\\\frac{dy}{dt}=-8\:\frac{m}{s}

We want to find the rate of change of the distance \frac{ds}{dt} when x = 9 and y = 12.

The distance of a point (x, y) and the origin is calculated by:

s=\sqrt{x^2+y^2}

We need to use the concept of implicit differentiation, we differentiate each side of an equation with two variables by treating one of the variables as a function of the other.

If we apply implicit differentiation in the formula of the distance we get

s=\sqrt{x^2+y^2}\\\\s^2=x^2+y^2\\\\2s\frac{ds}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}\\\\\frac{ds}{dt}=\frac{1}{s}(x\frac{dx}{dt}+y\frac{dy}{dt})

Substituting the values we know into the above formula

s=\sqrt{9^2+12^2}=15

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The rate of change of the distance \frac{ds}{dt} when x = 9 and y = 12 is -7.6\:\frac{m}{s}

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