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3 years ago

Which solution finds the value of x in the triangle below?

Mathematics

2 answers:

LuckyWell [14K]3 years ago

7 0

Answer:

Step-by-step explanation:

Since this is a right triangle, and one of the angles measures 60 degrees, we can conclude that the last side measures 30 degrees.

We can see that this is a 30-60-90 degree triangle.

The rules of 30-60-90 degree triangles are that the side opposite the 90 degree angle, or the hypotenuse can be measured with the variable $2a$ . The side opposite the 30 degree angle can be measured with $a$ , and the side opposite the 60 degree angle will be measured with $a\sqrt{3}$ .

We can see that 8 represents $2a$ because it is the hypotenuse. Since the side marked $x$ is separated by the hypotenuse by an angle of 60 degrees, we note that side marked $x$ is opposite the angle measuring 30 degrees. We note that the side opposite 30 degrees is marked $a$ , and since we already know that 8 is equal to $2a$ , we realize that the side marked x is equal to $a$ , or 4.

forsale [732]3 years ago

5 0

I didn't get it can you post a picture with it too?

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A very slow man walks 20 cm.in 0.5 left, what his velocity in meter per second?

Gre4nikov [31]

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2 years ago

A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 31 ft/s. (a) A

julsineya [31]

Answer:

a) -13.9 ft/s

b) 13.9 ft/s

Step-by-step explanation:

a) The rate of his distance from the second base when he is halfway to first base can be found by differentiating the following Pythagorean theorem equation respect t:

$D^{2} = (90 - x)^{2} + 90^{2}$ (1)

$\frac{d(D^{2})}{dt} = \frac{d(90 - x)^{2} + 90^{2})}{dt}$

$2D\frac{d(D)}{dt} = \frac{d((90 - x)^{2})}{dt}$

$D\frac{d(D)}{dt} = -(90 - x) \frac{dx}{dt}$ (2)

Since:

$D = \sqrt{(90 -x)^{2} + 90^{2}}$

When x = 45 (the batter is halfway to first base), D is:

$D = \sqrt{(90 - 45)^{2} + 90^{2}} = 100. 62$

Now, by introducing D = 100.62, x = 45 and dx/dt = 31 into equation (2) we have:

$100.62 \frac{d(D)}{dt} = -(90 - 45)*31$

$\frac{d(D)}{dt} = -\frac{(90 - 45)*31}{100.62} = -13.9 ft/s$

Hence, the rate of his distance from second base decreasing when he is halfway to first base is -13.9 ft/s.

b) The rate of his distance from third base increasing at the same moment is given by differentiating the folowing Pythagorean theorem equation respect t:

$D^{2} = 90^{2} + x^{2}$