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3 years ago

What is the exact length of AB? 4 m 8 mm m 2 TT m

Mathematics

1 answer:

dalvyx [7]3 years ago

5 0

Answer:

The answer is

$\pi$

if I am not wrong

Step-by-step explanation:

First you need to convert 45 degress to radians,then you use the arc length formula which is = thr radius * the angle to get the answer

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A notebook has a perimeter of 36 inches and an area of 80 square inches. What are the dimensions of the notebook?

maw [93]

Let L be length and W be width
The perimeter of a rectangle is 2W+2L=36
Thus 2W=36-2L, W=18-L
Area of a rectangle is WL=80
Thus (18-L)(L)=80
18L-L^2=80
L^2-18L+80=0
(L-8)(L-10)
Thus L can either be 8 or 10.
Because WL=80
If L is 8, then W=10
If L is 10, then W=8
So either way, the notebook is 8 by 10

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4 years ago

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Allisa [31]

Answer:

Step-by-step explanation:

inverse property of addition is going back to zero with the equation so adding 2 to negative 2 will go back to zero

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2 years ago

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Please can anyone help me with this I tried to do it and I got it wrong.

MatroZZZ [7]

Answer:

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Step-by-step explanation:

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3 years ago

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A recipe calls for 4 carrots for every 2 cups of water. Which expression can be used to find the number of carrots needed

Arada [10]

Answer:

4 divided by 2 = 2 so 2carrots per cup of water

Step-by-step explanation:

Hope this helps

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3 years ago

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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