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3 years ago

Help with this Math problem pleas?

Mathematics

1 answer:

victus00 [196]3 years ago

8 0

This a pretty typical right triangle trig problem; the first step is to figure out what we have and what we want in relation to an acute angle in the problem.

Here we have a right triangle, G=90°, and we're given angle F=23°. So we have to name everything in relation to F.

31 = FG is <em>adjacent </em>to F.

x = GE is <em>opposite </em>to F.

OK, we have an opposite and adjacent; that tells us we need to use the tangent of F. Let's write it:

tan 23° = tan F = opp/adj = x/31

Solving,

x = 31 tan 23°

I hate the calculator part. I used to love that part.

x = 31 tan 23° ≈ 13.16 feet

Answer: 4) x ≈ 13.2 ft

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John paid $300 for a new xbox with a discount of 40% what was the original price before discount

irakobra [83]

Answer:

$500 is the original price

Step-by-step explanation:

0.6x = 300 Divide 0.6 on both sides

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3 years ago

Read 2 more answers

The corners of a square lie on a circle of diameter d = 0.35 m. each side of the square has a length l. find l.

sergeinik [125]

Refer to the attached image.

As we can observe ABCD is a square, whose corners that is A,B,C and D lie on the circle.

The diameter AC = 0.35 meter.

We have to find the side of square.

Let the measure of side of a square be 'l' meter

In triangle ABC,

By Pythagoras theorem, we get

$(AC)^2 = (AB)^2 + (BC)^2$

$(0.35)^2 = (l)^2 + (l)^2$

$(0.35)^2 = 2(l)^2$

$\frac{(0.35)^2}{2} = (l)^2$

$l^2 = 0.06125$

So, l = 0.25 meter

Therefore, the measure of side of the square is 0.25 meter.

6 0

3 years ago

Two cars simultaneously left Points A and B and headed towards each other, and met after 2 hours and 45 minutes. The distance be

zheka24 [161]

<h2>Hello!</h2>

The answer is:

$FirstCarSpeed=41mph\\SecondCarSpeed=55mph$

To calculate the speed of the cars, we need to write two equations in order to create a relation between the two speeds and be able to isolate one in function of the other.

So, let be the first car speed "x" and the second car speed "y", writing the equations we have:

For the first car:

$x_{FirstCar}=x_o+v*t$

For the second car:

We know that the speed of the second car is the speed of the first car plus 14 mph, so:

$x_{SecondCar}=x_o+(v+14mph)*t$

Now, we already know that both cars met after 2 hours and 45 minutes, meaning that positions will be the same at that moment, and the distance between A and B is 264 miles, so, we can calculate the relative speed between them:

If the cars are moving towards each other the relative speed will be:

$RelativeSpeed=FirstCarSpeed-(-SecondCarspeed)\\\\RelativeSpeed=x-(-x-14mph)=2x+14mph$