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natulia [17]
3 years ago
10

PLEASE HELP !! ILL GIVE BRAINLIEST *EXTRA 40 POINTS* DONT SKIP :(( .!

Mathematics
1 answer:
Harrizon [31]3 years ago
3 0
180-53 is 127 angle 6 and 3 would be the same so your answer is 127
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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
A volleyball reaches its maximum height of 13 feet, 3 seconds after its served. Which of the following quadratics could model th
julsineya [31]
Only selections B and D give a maximum height of 13 at t=3. However, both of those functions have the height be -5 at t=0, meaning the ball was served from 5 ft below ground. This does not seem like an appropriate model.

We suspect ...
• the "correct" answers are probably B and D
• whoever wrote the problem wasn't paying attention.

8 0
3 years ago
Read 2 more answers
Help please I hate math
Hitman42 [59]

Answer:

Step-by-step explanation:

7 0
3 years ago
The length of a swimming pool is 3 feet longer than it’s width. The swimming pool is surrounded by a deck that is 2 feet wide an
Firlakuza [10]

Answer:

The Length of swimming pool is 8.35 feet

The width of swimming pool is 5.35 feet

Step-by-step explanation:

Given as :

The length of swimming pool is 3 feet longer that its width

Let The width of swimming pool = w  feet

So, The Length of swimming pool = L = (w + 3) feet

Now, The swimming pool is surrounded by deck of 2 feet wide

The width of deck = w' = w + 2 + 2 = (w + 4) feet

The area of deck = 116 feet²

The length of deck = L' = L + 2 + 2 = (L + 4) feet

So, L' = (w + 3 + 4) feet

I.e L' = (w + 7) feet

∵ The area of deck = 116 feet²

So , L' × w' = 116 feet²

(w + 7)× (w + 4) = 116

Or, w² + 4 w + 7 w + 28 = 116

Or, w² + 11 w - 88 = 0

Solving the quadratic equation as

ax² + bx + c = 0

So, w = \dfrac{-b\pm \sqrt{b^{2}-4\times a\times c}}{2\times a}

Or, w = \dfrac{-11\pm \sqrt{11^{2}-4\times 1\times (-88)}}{2\times 1}

Or, w = \dfrac{-11\pm \sqrt{473}}{2}

or, w = \dfrac{-11\pm 21.7}{2}

or, w = 5.35 , -16.35

The width of swimming pool = w = 5.35 feet

The Length of swimming pool = L = (5.35 + 3) feet = 8.35 feet

Hence, The Length of swimming pool is 8.35 feet

And The width of swimming pool is 5.35 feet

Answer

3 0
3 years ago
I need help on math on how to solve a linear equation. So how do you solve (x-2)+(2x+5).
saveliy_v [14]
So, put your "x's" together first and then your intercepts. That should answer your question :)
4 0
3 years ago
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