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kati45 [8]
2 years ago
11

How do you use a protractor

Mathematics
1 answer:
Vitek1552 [10]2 years ago
3 0

Place the midpoint of the protractor on the VERTEX of the angle.

Line up one side of the angle with the zero line of the protractor (where you see the number 0).

Read the degrees where the other side crosses the number scale.

hoped it helped :)

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Please help me with Part C of this question!!!
anzhelika [568]

Answer:

  a. 3/4 inches per minute

  b. -1 1/8 inches per minute

  c. B is fastest; 1 1/8 is more than 3/4

Step-by-step explanation:

A <em>change</em> is a <em>difference</em>. A <em>rate of change</em> is <em>one difference divided by another</em>, usually the change in y-value divided by the change in x-value.

__

<h3>a.</h3>

The change in elevation is the difference between the elevation at the end of the period (6 inches) and the elevation at the beginning of the period (3 inches). The change in time period is the difference between the end time (8 min) and the beginning time (4 min).

  change in elevation per minute = (6 -3 inches)/(8 -4 min)

  = (3 inches)/(4 min) = 3/4 inches/minute

__

<h3>b.</h3>

Similarly, ...

  change in elevation per minute = (3 -7 1/2 inches)/(18 -14 min)

  = (-4 1/2 inches)/(4 min) = -1 1/8 inches/minute

__

<h3>c.</h3>

We know that 3/4 is more than -1 1/8, but when we talk about the "fastest rate of change", we're generally interested in the magnitude--the value without the sign. That means we understand a rate of change of -1 1/8 inches per minute to be "faster" than a rate of change of 3/4 inches per minute.

The rate of change from Part B is fastest. 1 1/8 inches per minute is more than 3/4 inches per minute.

6 0
2 years ago
While sitting on a rooftop 95 feet off the ground Mary flicks a twig up and off of the ledge with the initial vertical velocity
cestrela7 [59]

Answer:

The maximum height of the twig is 96.6 feet off the ground

Step-by-step explanation:

To determine the maximum height of the twig,

First, we will determine the height distance covered by the twig after Mary flicks the twig up.

From the question,

Mary flicks a twig up and off of the ledge with the initial vertical velocity of 10 feet per second, that is

the initial velocity of the twig is 10 feet/second

At maximum height, the final velocity is 0

From on the equations of motions for bodies moving upwards,

v² = u² - 2gh

Where v is the final velocity

u is the initial velocity

g is the acceleration due to gravity (Take g = 9.8m/s² = 32.17 ft/s²)

and h is the height

From the question

u = 10 feet/second

and v = 0 feet/second

Putting the values into the equation

v² = u² - 2gh

0² = 10² - 2(32.17)h

0 = 100 - 64.34h

64.34h = 100

h = 100/64.34

h = 1.6 feet

This is the height distance covered by the twig after Mary flicks it up.

Now, the maximum height of the twig will be the sum of the height of the rooftop from the ground and the height distance covered by the twig.

That is,

Maximum height = 95 feet + 1.6 feet

Maximum height = 96.6 feet

Hence, the maximum height of the twig is 96.6 feet off the ground.

7 0
3 years ago
6th grade
Alex Ar [27]

Answer:

46

Step-by-step explanation:

69/3 = 23 (1/3 of the customers)

23×2 = 46 (2/3 of the customers)

3 0
3 years ago
I need to know how to solve this
kirza4 [7]
No solution, the answer is no solution
5 0
3 years ago
20 points!!!!! The table represents a linear function
Darya [45]
I believe the answer is 5. I multiplied 5 to -4 and that’s -20 and I’m guessing since slope-intercept form is y=Mx+b I’m thinking b=4 and the slope being 5 matches the output. -16=5(-4)+4
4 0
3 years ago
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