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Bess [88]
3 years ago
5

PLS HELP ILL MARK BRAINLIEST

Mathematics
2 answers:
ki77a [65]3 years ago
5 0

Answer:

(-4,-6),(-2,-5),(0,-4),(4,-2)

Step-by-step explanation:

(x,y)

x represents the x co-ordinate of a point

y represents the y co-ordinate of a point

aleksandrvk [35]3 years ago
4 0

Answer:

-6,-2,0,4

Step-by-step explanation:

Since x comes before y in an ordered pair, you would need to find the where the x is, then search for the y number it connects with. For finding the x, use the y part, and look for the numbe it connects with on the graph. In this case youy pairs would be, (-4,-6) (-2,-5) (0,-4) (4,-2). Hope I helped!

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An air traffic controller spots two airplanes at the same altitude converging to a point as they fly at right angles to each oth
sattari [20]
They are traveling at right angles to each other so we can say one is traveling north to south and the other west to east.  Then we can say that there positions, y and x are:

y=150-600t  x=200-800t

By using the Pythagorean Theorem we can find the distance between these two planes as a function of time:

d^2=y^2+x^2, using y and x from above

d^2=(150-600t)^2+(200-800t)^2

d^2=22500-180000t+360000t^2+40000-320000t+640000t^2

d^2=1000000t^2-500000t+62500

d=√(1000000t^2-500000t+6250)

So the rate of change is the derivative of d

dd/dt=(1/2)(2000000t-500000)/√(1000000t^2-500000t+6250)

dd/dt=(1000000t-250000)/√(1000000t^2-500000t+6250)

So the rate depends upon t and is not a constant, so for the instantaneous rate you would plug in a specific value of t...

...

To find how much time the controller has to change the airplanes flight path, we only need to solve for when d=0, or even d^2=0...

1000000t^2-500000t+62500=0

6250(16t^2-8t+1)=0

6250(16^2-4t-4t+1)=0

6250(4t(4t-1)-1(4t-1))=0

6250(4t-1)(4t-1)=0

6250(4t-1)^2=0

4t-1=0

4t=1

t=1/4 hr

Well technically, the controller has t<1/4 because at t=1/4 impact will occur :)


7 0
3 years ago
The kittens, Annie and Josie, are pushing a ball, Annie with a force magnitude of 80 N in a direction of 133 degrees, and Josie
kvv77 [185]

Answer: Then the magnitude of the force is 37.86N, and the direction is 54.35°

Step-by-step explanation:

We can write the forces as vectors.

We know that Annie pushes with a magnitude of 80N in a direction of 133° (Remember that the angles are always measured from the x-axis)

The components of this force, (Ax, Ay), are then:

Ax = 80N*cos(133°)

Ay = 80N*sin(133°)

And we know that Josie pushes with a magnitude of 95N in direction of 290°

The components of this force, (Jx, Jy), are:

Jx = 95N*cos(290°)

Jy = 95N*sin(290°)

When we add these forces, the total force acting on the ball is:

F = (80N*cos(133°) ,  80N*sin(133°)) + (95N*cos(290°), 95N*sin(290°))

F = (80N*cos(133°) + 95N*cos(290°), 80N*sin(133°) + 95N*sin(290°))

Now, the third kitten wants to do a force K, in a direction θ, such that the net force acting on the ball is zero.

Then we must have that, each component of the force of the third cat (K*cos(θ)  on the x-axis and k*sin(θ) on the y-axis), is such that:

K*cos(θ) + 80N*cos(133°) + 95N*cos(290°) = 0

k*sin(θ)  + 80N*sin(133°) + 95N*sin(290°) = 0

Now we need to solve that system for k and θ

if we simplify the equations we get:

k*cos(θ) - 22.07N = 0

k*sin(θ)  -30.76N  = 0

Now we can rewrite them as:

k*cos(θ) = 22.07N

k*sin(θ)   = 30.76N  

Now we can take the quotiet between both equations to get:

(k*sin(θ))/(k*cos(θ)) = 30.76N/22.07N

Tan(θ) = 1.394

θ = Atan(1.394) = 54.35°

Now that we know the angle, we can find the value of the magnitude k, by using one of the two equations of the system:

k*cos(54.35°) = 22.07N

k = 22.07N/cos(54.35°) = 37.86N

Then the magnitude of this force is 37.86N, and the direction is 54.35°

7 0
3 years ago
BRAINLIEST<br>x + 4 &gt; 8. If you need to solve for x, what's the first step?
Darina [25.2K]

Answer:

the first step is subtract 4 from both sides

x + 4 > 8

x + 4 - 4 > 8 - 4

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Which statement is false?
Sidana [21]

Trigonometric functions which are related by having the same value at complementary angles are called cofunctions. Cofunctions of complementary angles are equal.

A. csc 20' = csc(90-70)=sec 70

B. cos 87' =  cos (90-3)=sin 3'

C. csc 40' =  csc(90-50) =sec50'  


D. tan 15' = tan(90-75)= cot 75'

Among all the option  c is not correct.

Option C is false.




8 0
4 years ago
Read 2 more answers
What is the pattern 3.5 to 13
mars1129 [50]
Adding 9.5 every time

4 0
3 years ago
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