Answer:
<h2>9/6</h2>
Step-by-step explanation:
Remember that, to find the reciprocal of a fraction, just swap the numbers. So the reciprocal of 6/9 is 9/6.
Answer:

Step-by-step explanation:
<u>Step 1: Substitute -3x for x</u>
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Answer: 
Answer:
C
Step-by-step explanation:
1)Non- zeros (1,2,3,....) are always significant.
2)Captive zeros are significant too( zeros between non- zeros ).
3)Trailing zeros (zeros at the end) aren't significant always unless if there is a decimal point.
Such this question we have decimal point so we count them
Answer:
Length = 16 inches and width = 12inches
Step-by-step explanation:
A = 192 in² . The width is 4 + 1/2x. therefore let x = length
Area = LW
192 = x(4 + 1/2 x)
192 = 4x + 1/2x²
384 = 8x + x² Multiply each term by 2 to make a
384 + 16 = x² + 8x + 16 Complete the square
400 = (x + 4)²

± 20 = x + 4
20 = x + 4 or -20 = x + 4
20 - 4 = x or -20 - 4 = x
16 = x or -24 = x reject the negative amount
Length = 16
width = 4 + 1/2(16); 4 + 8 = 12
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:

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

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\\\\](https://tex.z-dn.net/?f=%5Cdfrac%7Bdt%7D%7Bdx%7D%3D%5Cdfrac%7B1%7D%7B4%7D%2B%5Cdfrac%7B1%7D%7B1.1%7D%28%5Cdfrac%7B1%7D%7B2%7D%29%5Cdfrac%7B2x-120%7D%7B%5Csqrt%7Bx%5E2-120x%2B5200%7D%7D%20%5C%5C%5C%5C%5C%5C%5Cdfrac%7Bdt%7D%7Bdx%7D%3D%5Cdfrac%7B1%7D%7B4%7D%20%2B%5Cdfrac%7B1%7D%7B1.1%7D%20%5Cdfrac%7Bx-60%7D%7B%5Csqrt%7Bx%5E2-120x%2B5200%7D%7D%20%3D0%5C%5C%5C%5C%5C%5C%20%20%5Cdfrac%7Bx-60%7D%7B%5Csqrt%7Bx%5E2-120x%2B5200%7D%7D%20%3D%5Cdfrac%7B1.1%7D%7B4%7D%3D%5Cdfrac%7B2%7D%7B7%7D%5C%5C%5C%5C%5C%5C%20x-60%3D%5Cdfrac%7B2%7D%7B7%7D%5Csqrt%7Bx%5E2-120x%2B5200%7D%5C%5C%5C%5C%5C%5C%28x-60%29%5E2%3D%5Cdfrac%7B2%5E2%7D%7B7%5E2%7D%28x%5E2-120x%2B5200%29%5C%5C%5C%5C%5C%5C%28x-60%29%5E2%3D%5Cdfrac%7B4%7D%7B49%7D%5B%28x-60%29%5E2%2B40%5E2%5D%5C%5C%5C%5C%5C%5C%281-4%2F49%29%28x-60%29%5E2%3D4%2A40%5E2%2F49%3D6400%2F49%5C%5C%5C%5C%2845%2F49%29%28x-60%29%5E2%3D6400%2F49%5C%5C%5C%5C45%28x-60%29%5E2%3D6400%5C%5C%5C%5C)

As
, 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.