I think the answer is six
(1,1) , (2,2) , (3,3) , (4,4) , (5,5) , (6,6)
Answer:
the link dosent work says it have bug in it
Step-by-step explanation:
<span>Find the wind speed and the plane's airspeed.
:
Let s = speed of the plane in still air
Let w = speed of the wind
then
(s-w) = plane speed against the wind
and
(s+w) = plane speed with the wind
:
Change 3 3/8 hrs to 3.375 hrs
:
The trips there and back are equal distance, (1890 mi) write two distance equations
dist = time * speed
:
3.375(s-w) = 1890
3.0(s + w) = 1890
:
It is convenient that we can simplify both these equations:
divide the 1st by 3.375
divide the 2nd by 3
resulting in two simple equations that can be used for elimination of w
s - w = 560
s + w = 630
----------------adding eliminates w, find s
2s = 1190
s =
s = 595 mph is the plane speed in still air
Find w
595 + w = 630
w = 630 - 595
w = 35 mph is the wind spee</span>
The terminal point represented by P(x,y), is the coordinate where a certain point lies. Given is 5pi/4 radians - this will be our angle.
Evaluate the values of x and y using sin θ and cos θ
x = cos (5pi/4)
x = -√2/2
y = sin (5pi/4)
y = -√2/2
Therefore, (x,y) is (-√2/2, -√2/2)
This is a GCF problem,
find the GCF of 36 and 27. To do this, list out the factors for each number.
EX 36- 1×36, 2×18, so on and so on. Then do this for 27. The GCF will be the greatest factor. In this case, that is 9.
So there would be 9 groups because that is the greatest common factor.
there would be 4 roses in each group because 4×9= 36.
There are 3 carnations in each group because 9×3 =27.
So 9 groups with 4 roses and 3 carnations.
