The speed of wind and plane are 105 kmph and 15 kmph respectively.
<u>Solution:</u>
Given, it takes 6 hours for a plane to travel 720 km with a tail wind and 8 hours to make the return trip with a head wind.
We have to find the air speed of the plane and speed of the wind.
Now, let the speed wind be "a" and speed of aeroplane be "b"
And, we know that, distance = speed x time.
Now at head wind → 
So, solve (1) and (2) by addition
2a = 210
a = 105
substitute a value in (1) ⇒ 105 + b = 120
⇒ b = 120 – 105 ⇒ b = 15.
Here, relative speed of plane during tail wind is 120 kmph and during head wind is 90 kmph.
Hence, speed of wind and plane are 105 kmph and 15 kmph respectively.
Abd is a right angled triangle so by pythagoras we know
and to calculate h we need to use the sine rule
i hope this helps
What are you asking here? I need more information to solve please
Stop posting on here and do the work yourself I usually don't mind helping people but in some cases, I need to just say do it by yourself look at notes look up the answer there are a million things you can do instead of spam posting all your questions good day
Answer:
(3x+4)(5x+7)
Step-by-step explanation:
15x^2
+41x+28
Factor the expression by grouping. First, the expression needs to be rewritten as 15x^2
+ax+bx+28. To find a and b, set up a system to be solved.
a+b=41
ab=15×28=420
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 420.
1,420
2,210
3,140
4,105
5,84
6,70
7,60
10,42
12,35
14,30
15,28
20,21
Calculate the sum for each pair.
1+420=421
2+210=212
3+140=143
4+105=109
5+84=89
6+70=76
7+60=67
10+42=52
12+35=47
14+30=44
15+28=43
20+21=41
The solution is the pair that gives sum 41.
a=20
b=21
Rewrite 15x^2
+41x+28 as (15x^2
+20x)+(21x+28).
(15x^2
+20x)+(21x+28)
Factor out 5x in the first and 7 in the second group.
5x(3x+4)+7(3x+4)
Factor out common term 3x+4 by using distributive property.
(3x+4)(5x+7)
