Anyone know this answer

alexandr1967 [171]

Remember that f(a) is the same thing as y just about. So the y value is -2. That's the flat part of the graph about.

You need an x value that is along the flat part of the graph.

The flat part goes from x = - 1 to x = 0

You need a value between 0 and - 1. The only value there that does that is B.
I don't know why the answer was given to you as A. It is not correct. Sometimes questions are like that. There's just an error in them. Appeal it with your teacher.

B <<<<<< answer.

6 0

3 years ago

Write a 5 paragraph essay about the Three types of symmetry

Hoochie [10]

B r u h h h h h h h hh h h

8 0

3 years ago

Read 2 more answers

Two forces with magnitudes of 100 and 50 pounds act on an object at angles of 50° and 160°, respectively. Find the direction and

MatroZZZ [7]

<h2>Hello!</h2>

Since there is no additional information, we can assume that the given angles are respect to the horizontal axis (x), so

The answer is:

Magnitude: 95.14 lbs

Direction: 80° respect to the x-axis

Since there is no additional information about the given angles, we can assume that both angles are respect to the horizontal axis (x), positive direction to x-axis positive directions (right) and above x-axis.

<h2>Calculations:</h2>

100 pounds (lb) force, 50° angle:

$100lbs*cos(50)=64.28lbs\\100lbs*sin(50)=76.60lbs$

50 pounds (lb) force, 160° angle:

$50lbs*cos(160)=-46.98lbs\\50lbs*sin(160)=17.10lbs$

So,

For x-axis forces we have:

64.28lbs (positive direction)

-46.98lbs (negative direction)

Then,

$64.28lbs-46.98lbs=17.3lbs$ (positive direction)

Fox y-axis forces we have:

76.60lbs (positive direction)

17.10 lbs (positive direction)

Then,

$76.60lbs + 17.10 lbs =93.7lbs$ (positive direction)

So, we have the two components values of the acting force:

17.3 lbs for the x-axis and 93.7 lbs for the y-axis.

Therefore,

We have to calculate the angle between both resultant force components:

Let's calll it α

So,

$\alpha=tan^{-1}(\frac{y-axisValue}{x-axisValue})=tan^{-1}(\frac{93.7}{17.3})\\\alpha=79.5=80$

Hence, the angle between both resultant force components is 80 °, meaning that the resultant force direction is positive, above the x-axis.

Now we have to calculate the magnitude of the resultant force since the sum of both resultant force components is equal to the hypotenuse of the formed triangle, we have that:

$sin(\alpha)=\frac{y-axisValue}{ResultantForceMagnitude}=\frac{93.7lbs}{ResultantForceMagnitude}\\\\ResultantForceMagnitude=\frac{93.7lbs}{sin(80)}=95.14lbs$