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3 years ago

6. In Diagram 1, NPQRSTUV is a regular octagon. NVW and STW are straight lines.

Mathematics

1 answer:

Sveta_85 [38]3 years ago

3 0

Answer:

A. 37.5°

Step-by-step explanation:

∠SPT can be considered an inscribed angle in the circumcircle. In that circle, arc ST would have a measure of 360°/8 = 45°. Angle y° is half that measure, so is 22.5°.

The external angles of a regular octagon are 360°/8 = 45°, so that is the measure of angles UTW and UVW. It is also the amount by which reflex angle TUV exceeds 180°.

The sum of the interior angles of quadrilateral TUVW is ...

45° +225° +45° +3x° = 360°

Then 3x° = 45° and x° = 15°.

Then the sum of interest is ...

x° + y° = 15° + 22.5° = 37.5°

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2 years ago

Francisco needed to get his computer fixed. He took it to the repair

grin007 [14]

Answer:

$95 per hour

Step-by-step explanation:

The total of charges is the sum of the cost of labor and the cost of parts. The cost of labor is the product of hours (5) and the cost per hour (x). Then the equation for total cost is ...

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2 years ago

A ball is thrown straight out at 80 feet per second from an upstairs window that's 15 feet off the ground. Find the ball's horiz

Alex73 [517]

Answer:

Step-by-step explanation:

In order to find the horizontal distance the ball travels, we need to know first how long it took to hit the ground. We will find that time in the y-dimension, and then use that time in the x-dimension, which is the dimension in question when we talk about horizontal distance. Here's what we know in the y-dimension:

a = -32 ft/s/s

v₀ = 0 (since the ball is being thrown straight out the window, the angle is 0 degrees, which translates to no upwards velocity at all)

Δx = -15 feet (negative because the ball lands 15 feet below the point from which it drops)

t = ?? sec.

The equation we will use is the one for displacement:

Δx = $v_0t+\frac{1}{2}at^2$ and filling in:

$-15=(0)t+\frac{1}{2}(-32)t^2$ which simplifies down to

$-15=-16t^2$ so

$t=\sqrt{\frac{-15}{-16} }$ so

t = .968 sec (That is not the correct number of sig fig's but if I use the correct number, the answer doesn't come out to be one of the choices given. So I deviate from the rules a bit here out of necessity.)

Now we use that time in the x-dimension. Here's what we know in that dimension specifically:

a = 0 (acceleration in this dimension is always 0)

v₀ = 80 ft/sec

t = .968 sec

Δx = ?? feet

We use the equation for displacement again, and filling in what we know in this dimension:

Δx = $(80)(.968) +(0)(.968)^2$ and of course the portion of that after the plus sign goes to 0, leaving us with simply: