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dybincka [34]
1 year ago
10

The path of a race will be drawn on a coordinate grid like the one shown below. The starting point of the race will be at (−5.4,

3). The finishing point will be at (3, −5.4). A four-quadrant coordinate grid is drawn from negative 4 to positive 4. Quadrant I is labeled Quadrant P, Quadrant II is labeled Quadrant Q, Quadrant III is labeled Quadrant R, and Quadrant IV is labeled Quadrant S. Part A: Use the grid to determine in which quadrants the starting point and the finishing point are located. Explain how you determined the locations. (6 points) Part B: A checkpoint will be at (5.4, 3). In at least two sentences, describe the difference between the coordinates of the starting point and the checkpoint, and explain how the points are related. (4 points)BEST ANSWER GETS BRAINLIEST
Mathematics
1 answer:
IceJOKER [234]1 year ago
8 0

Answer:

SP = Quadrant 3    FP = Quadrant  4

Step-by-step explanation:

sp =  (−5.4, 3) fp (3, −5.4)  this means the line is decreasing as point y coordinate  changes fro 3 to 5.4 and will be a negative slope gradient- The quadrants at top left to right are 3 and 1  and bottom left to right is 2 and 4  Labels are Quadrant 3 = R      Quadrant 1 = P  Quadrant 2 = Q  Quadrant 4 = S The starting point is Left top side as y is positive and can only be 3 or 1 and x is negative      SP = Quadrant 3    The finishing point is bottom quadrants as y is negative this time and x is positive      FP = Quadrant  4   You can remember that it is often start of origin then quadrant 1 in straight line positive colorations

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Find a single expression that represents the area of the outer ring of the circle if the area of the whole circle is represented
sp2606 [1]

Answer:  The answer is A_o=2\pi\sqrt{\dfrac{7}{22}(25x^2-12x-9)}.


Step-by-step explanation: Given that the area of the whole circle is represented by the expression

A_c=25x^2-12x-9.

We are to find the area of the outer ring of the circle, i.e., to find the circumference of the circle.

Now, if 'r' represents the radius of the circle, then we have

A_c=\pi r^2\\\\\Rightarrow \dfrac{22}{7}r^2=25x^2-12x-9\\\\ \Rightarrow r^2=\dfrac{7}{22}(25x^2-12x-9)\\\\\Rightarrow r=\sqrt{\dfrac{7}{22}(25x^2-12x-9)}.

Thus, the area of the outer ring is

A_o=2\pi r=2\pi\sqrt{\dfrac{7}{22}(25x^2-12x-9)}.


5 0
2 years ago
A campus deli serves 300 customers over its busy lunch period from 11:30 a.m. to 1:30 p.m. A quick count of the number of custom
Irina-Kira [14]

Complete question is;

A campus deli serves 300 customers over its busy lunch period from 11:30 am to 1:30 pm. A quick count of the number of customers waiting in line and being served by the sandwich makers shows that an average of 10 customers are in process at any point in time. What is the average amount of time that a customer spends in process?

Answer:

4 minutes

Step-by-step explanation:

For this question, we will apply Little's law which is is a theorem that determines the average number of items in a stationary queuing system, based on the average waiting time of an item within a system and the average number of items arriving at the system per unit of time.

The formula for the law is:

Inventory = flow rate × flow time.

We are given;

Inventory = 300 customers

flow time is from 11: 30am to 1:30pm which is 2 hours = 120 minutes

Flow rate = 300/120 = 2.5 persons/minute

Now, Making flow rate the subject of the formula earlier given, we have;

flow rate =  inventory/ flow time

Flow time is the time each person spends in the process

Thus, plugging in the relevant values, we get ;

We are told that an average of 10 customers are in process at any point in time.

Thus;

Average flow time = average inventory/flow rate = 10/2.5 = 4 minutes

3 0
2 years ago
Write a real world problem that can be represented by the equation 3/4c=21
Black_prince [1.1K]

Answer: Real world problem is "A student have c toffee he distribute \frac{3}{4}th part of those toffees to his friends. He gave total 21 toffees to his friend".

Explanation:

Let a student have c number of toffees in his bag.

It is given that he distribute \frac{3}{4}th part of those toffees to his friends.

The \frac{3}{4}th part of c toffees is,

\frac{3c}{4}

The total number of distributed toffees is 21.

\frac{3c}{4}=21

It is the same as given equation.

If we change the equation in words it means the \frac{3}{4}th part of a number c is 21.

4 0
2 years ago
Read 2 more answers
Hey ^-^ can someone please help me with this problem:
soldi70 [24.7K]

Answer:   8 (Pi - sqrt(3))

Discussion:

The area of the shaded region is that of the semicircle minus the area of the triangle..

Area of semicircle = 1/2 * Pi * R^2        

   Where R^2 is the square of the radius of the circle. In our case, R ( = OC)

    = 4 so the semicircle area is

   (1/2) * Pi * (4^2) = (1/2) * Pi * 16 = 8 Pi

Area of triangle.

  First of all, angle ACB is a right angle ( i.e. 90 degrees).

    * This is the Theorem of Thales from elementary Plane Geometry. *

 so by Pythagoras

   AC^2 + BC^2 = AB^2

But CB = 4 (given) and AB = 4*2 = 8 ( the diameter is twice the radius).

Substituting these in Pythagoras gives

   AC^2 + 4^2 = 8^2 or

   AC^2 = 8^2 - 4^2- = 64 - 16 = 48

   Hence AC = sqrt(48) = sqrt (16*3) = 4 * sqrt(3)

We are almost done! The area of the triangle is given by

  (1/2) b * h = (1/2)  BC * AC = (1/2) 4 * (4 * sqrt(3)) =  8 sqrt(3)

We conclude the area area of the shaded part is

 8 PI - 8 sqrt(3)   = 8 (Pi - sqrt(3))

Note that sqrt(3) is approx  1.7 so (PI - sqrt(3)) is a positive number, as it better well be!

6 0
2 years ago
A bug was running along a number line at a speed of 11 units per minute. It never changed its direction. If at 7:15 pm it was at
Ne4ueva [31]
7:15-7:20 is 5 minutes
If every one minute is 11 units then it would be 5x11=55
If it was at point 100 it would now be at point 100+55=155
Answer: point 155
Hope this is correct and helps :)
3 0
2 years ago
Read 2 more answers
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