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2 years ago

A sprinter can run 279 feet in 9 seconds. find the sprinter's unit rate of feet per second.

Mathematics

2 answers:

kolezko [41]2 years ago

7 0

Answer: The required sprinter's unit rate is 31 feet per second.

Step-by-step explanation: Given that a sprinter can run 279 feet in 9 seconds.

We are to find the sprinter's unit rate of feet per second.

We will be using the UNITARY method to solve the given problem.

According to the given information, we have

In 9 second, distance run by the sprinter = 279 feet.

Therefore, in 1 second, the distance run by sprinter will be

$\dfrac{279}{9}=31~\textup{feet per second}.$

Thus, the required sprinter's unit rate is 31 feet per second.

MariettaO [177]2 years ago

3 0

279ft/9sec = xft/1sec
Cross multiply to get
9x = 279 divide by 9
x = 31
31 feet per second

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The center of a circle is at the origin on a coordinate grid. The vertex of a parabola that opens upward is at (0, 9). If the ci

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Answer:

"The maximum number of solutions is one."

Step-by-step explanation:

Hopefully the drawing helps visualize the problem.

The circle has a radius of 9 because the vertex is 9 units above the center of the circle.

The circle the parabola intersect only once and cannot intercept more than once.

The solution is "The maximum number of solutions is one."

Let's see if we can find an algebraic way:

The equation for the circle given as we know from the problem without further analysis is so far $x^2+y^2=r^2$ .

The equation for the parabola without further analysis is $y=ax^2+9$ .

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$x^2+(ax^2+9)^2=r^2$

To expand $(ax^2+9)^2$ , I'm going to use the following formula:

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$(ax^2+9)^2=a^2x^4+18ax^2+81$ .

$x^2+y^2=r^2$

$x^2+(ax^2+9)^2=r^2$

$x^2+a^2x^4+18ax^2+81=r^2$

So this is a quadratic in terms of $x^2$

Let's put everything to one side.

Subtract $r^2$ on both sides.

$x^2+a^2x^4+18ax^2+81-r^2=0$

Reorder in standard form in terms of x:

$a^2x^4+(18a+1)x^2+(81-r^2)=0$

The discriminant of the left hand side will tell us how many solutions we will have to the equation in terms of $x^2$ .

The discriminant is $B^2-4AC$ .

If you compare our equation to $Au^2+Bu+C$ , you should determine $A=a^2$

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$C=(81-r^2)$

The discriminant is

$B^2-4AC$

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Multiply the (18a+1)^2 out using the formula I mentioned earlier which was:

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We know that $r>0$ because in order it to be a circle a radius has to exist.

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We have to go further to see.

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$\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}$

This is t he solution for $x^2$ .

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$x=\pm \sqrt{\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}}$

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That is,

$x=\pm \sqrt{\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}}$

means you have:

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$x=\pm \sqrt{\frac{-(18a+1)-\sqrt{36a+1+4a^2r^2}}{2a^2}}$ .

The second one is definitely includes a negative result in the square root.

18a+1 is positive since a is positive so -(18a+1) is negative

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Also if (0,9) is intersection, then

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After this confirmation it would be interesting to see what happens with assume algebraically the solution should be (0,9).

This means we should have got x=0.

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A fraction is only 0 when it's top is 0.

$0=-(18a+1)+\sqrt{36a+1+4a^2r^2}$

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$324a^2=4a^2r^2$

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$9=r$

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Let's go through the radius choices.

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