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1 year ago

Which of the following equations can be used to find the length of EF in the

Mathematics

1 answer:

Vinil7 [7]1 year ago

3 0

The equation which can be used to find the length of EF is option A which is EF= $\sqrt{24^{2} -12^{2} }$ .

Given the length of base of right angled triangle is 12 and the hypotenuse be 24.

We are to find the length of EF and tell which equation can be used to find the length of EF.

We know that pythagoras theorem applies in a right angled triangle.

Pythagoras theorem says that in a right angled triangle the square of the hypotenuse is equal to the sum of square of base and perpendicular.

$H^{2} =B^{2} +P^{2}$

In this triangle we are only given base and hypotenuse and we have to find EF which is perpendicular.

$(DE)^{2} =(EF)^{2} +(DF)^{2}$

EF= $\sqrt{DE^{2}-DF^{2} }$

= $\sqrt{(24)^{2} -(12)^{2} }$ --1

= $\sqrt{576-144}$

= $\sqrt{432}$

Hence the equation which can be used to find the length of Ef is option A which is EF= $\sqrt{24^{2} -12^{2} }$ .

Question is incomplete as it should include the figure also.

Learn more about pythagoras theorem at brainly.com/question/343682

#SPJ1

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3 years ago

You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along

sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

$d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}$

Then, the time (speed divided by distance) is:

$t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1$

To optimize this function we have to derive and equal to zero:

$\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\ \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\$

$x$

As $d_b=x$ , the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.