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3 years ago

Heres another math problem

Mathematics

2 answers:

ser-zykov [4K]3 years ago

8 0

Answer:

50°

Step-by-step explanation:

Now look at the straight line below,We know that in a straight line sum of all angles is equal to 180°. so the remaining unknown angle in that line is 180°-150°=30° and that angle is equal to the angle in the upper straight line as they are vertically opposite angles in the intersection of two straight lines.

Now apply the same concept for the upper straight line also that is sum of the angles 30° and 50° and 2x° is equal to 180° .

30+50+2x=180.

->80+2x=180.

->2x=100

x=50°

MrRissso [65]3 years ago

4 0

Answer:

$\therefore x=50\°$

Step-by-step explanation:

Let the angle between $2x$ and 150° be = b°

$150\°+b\°=180\°$ [ Forming a linear pair of angles]

Solving for $b$ by subtracting both sides by 150.

$150+b-150=180-150$

∴ $b=30\°$

$50\°+2x+b\°=180\°\\$ [Angle lying on a straight line add up to 180]

Plugging in $b=30\°$

$50+2x+30=180\\2x+80=180$

Solving for $x$ by

subtracting both sides by 80.

$2x+80-80=180-80\\2x=100$

then dividing both sides by 2.

$\frac{2x}{2}=\frac{100}{2}\\\\\therefore x=50\°$

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Answer:

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Step-by-step explanation:

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3 years ago

A boat is pulled into a dock by means of a rope attached to a pulley on the dock. The rope is attached to the front of the boat,

san4es73 [151]

Answer:

The boat will be approaching the dock at 14.06 ft/min

Step-by-step explanation:

The very first thing we need to do when solving this kind of problems is to draw a diagram that will represent the problem. (Diagram attached).

When looking at the diagram, we can see that we can build a model of the situation by using a right triangle, whose height will be constant while its base (x) and its hypotenuse (l) are variable. These are the variables we are going to work with (Check uploaded picture of the triangle).

Once we have our triangle correctly set up, we can use the pythagorean theorem to relate each of the sides of the triangle.:

Pythagorean theorem: $c^{2}=a^{2}+b^{2}$

So the equation for our problem is:

$l^{2}=x^{2}+10^{2}$

Since we need to find the horizontal speed of the boat, we will need to solve the equation for x, so we get:

$x^{2} =l^{2}-10^{2}\\\\x=\sqrt{l^{2}-10^{2}}$

We can now use this equation to find the horizontal velocity of the boat. Position is related to velocity in that the derivative of position with respect to time is the instantaneous velocity of the parcticle at a given time t.

$\frac{dx}{dt}=velocity$

so we need to take the derivative of the given equation. We start by rewritting the radical as a rational power, so we get:

$x=(l^{2}-100)^{\frac{1}{2}}$

and now we can use the chain rule to take the derivative of this function. We take the derivative of the outern function and multiply it by the derivative of the inner function, like this:

$dx=\frac{1}{2}(l^{2}-100)^{-\frac{1}{2}}(2l)dl$

now we can simplify the equation and divide both sides of it into dt so we get:

$\frac{dx}{dt}=\frac{l}{\sqrt{l^{2}-10^{2}} }\frac{dl}{dt}$

the dx/dt part will represent the horizontal velocity of the boat, while the dl/dt part will represent the velocity at which the rope is being pulled, so we have enough data to solve our problem

$\frac{dx}{dt}=\frac{l}{\sqrt{l^{2}-10^{2}} }\frac{dl}{dt}\\\\\frac{dx}{dt}=\frac{110}{\sqrt{110^{2}-10^{2}} }*14\\\\\frac{dx}{dt}=\frac{110}{\sqrt{12000} }*14\\$