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3 years ago

In AKLM, m = 1.2 cm, k = 5.1 cm and

Mathematics

1 answer:

Elodia [21]3 years ago

4 0

Answer:

4.2 cm

Step-by-step explanation:

The law of cosines is applicable.

l² = k² +m² -2km·cos(L)

l² = 5.1² +1.2² -2·5.1·1.2·cos(35°) ≈ 17.4236

l ≈ √17.4236

l ≈ 4.2 . . . cm

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An open-topped glass aquarium with a square base is designed to hold 444 cubic feet of water. What is the minimum exterior surfa

In-s [12.5K]

Answer:

291.05ft²

Step-by-step explanation:

Volume of the aquarium = L³

L is the length of one side of the aquarium

444 = L³

L = cuberoot[444]

L= 7.63ft

Since the glass aquarium is opened at the top

Total surface Area = 5L²

Total surface Area = 5(7.63)²

Total surface Area = 291.05ft²

Hence the minimum exterior surface area of the aquarium is 291.05ft²

8 0

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.

7 0

3 years ago

Which graph can be used to find the solution(s) to 4x=-4x?

Elden [556K]

<h2>Answer:</h2>

The graph is shown in the Figure below

<h2>Step-by-step explanation:</h2>

In this exercise, we have an equation. On the left side we have a straight line with slope $m=4$ and there is no any y-intercept. On the right side, on the other had, we also have a straight line, but the slope here is $m=-4$ . Therefore, by plotting these two straight lines, we have that the solution is the origin, that is, the point $(0, 0)$ .