can u help me?

Question

I NEED ANSWERS
TWO GOODS TRAINS LEAVE A RAILWAY STATION AT THE SAME TIME. THE FIRST TRAIN TRAVELS DUE WEST AND THE SECOND DUE NORTH. THE FIRST TRAIN TRAVELS 5KM/HR FASTER THAN THE SECOND TRAIN. IF AFTER 2 HRS. THEY R 50KM APART FIND THE AVERAGE SPEED OF EACH TRAIN.

2. A PERSON SELLS 2 ARTICLES FOR RS.46 MAKING A PROFIT OF 10% ON ONE AND 20% ON THE OTHER. IF HE HAD SOLD EACH ARTICLE AT 15% PROFIT THE RESULT IS SAME. AT WHAT PRICE DOES HE SELL EACH ARTICLE?

THE ABOVE QUE ARE LINEAR EQUATIONS IN TWO VARIABLES.

TRIGONOMETRY QUESTIONS

3. FROM A WINDOW 'H' METRES HIGH ABOVE THE GROUND OF A HOUSE IN A STREET THE ANGLES OF ELEVATION AND DEPRESSION OF THE TOP AND THE FOOT OF ANOTHER HOUSE ON THE OPPOSITE SIDE OF THE STREET ARE 'DETA' AND 'FINI' RESPECTIVELY. SHOW THAT HEIGHT OF OPPOSITE HOUSE IS H{1+TAN DETA * COT FINI}

PLS NOTE THAT DETA AND FINI ARE ANGLES. PLS HELP

EVEN I AM ON SUMMER BREAK PLS HELP AND I HAVE TRIED 100 TIMES BUT SILL NOT GETTING IT.

TURAS WHICH R U ANSWERING. THANK U SO MUCH FOR THE FIRST QUESTION.

Answer 1

do your own homework dude

Answer 2

Let the first train travel at a speed of x kmph and the second at y kmph. Then x - y = 5, given x travels at a speed 5 kmph higher than y
after 2 hrs first train would have travelled 2xkm and the second train would have travelled 2y km and since they; are moving perpendicular to each other, i.e west and north, the distance between them will be (2x)^2 + (2y)^2 and this will be equal to
(50)^2 and hence, (2x)^2 + (2y)^2 = 2500 and solving the two equations we get, x = 20 and y = 15. Hence the first train travels at a speed of 20 kmph and the second train travels at the speed of 15 kmph.

second problem:

Let the cost of one article be x and the other y.

then x + y = 46

by geting profit of 10% on article x and 20% on article y we get total profit as 0.1x + 0.2y and making a profit of 15% on each article the total profit works out to 0.15x + 0.15y Sin;ce the two profits are equal we have

0.1x + 0.2y = 0.15x + 0.15y or multiplying both sides by 100 we have, 10x + 20y = 15x + 15y or 5x - 5y = 0 or x - y = 0 and we have x + y = 46 and solving the 2 equations we have
x = 23 and y = 23 and hece he sells each article at 23.

Answer 3

The first question:
The first train is traveling at (r+5) for 2 hours for a distance of 2r+10.
The second train is traveling at r for 2 hours for a distance of 2r. Using the Pythagorean Theorem: (2r)^2 + (2r+10)^2 = 50^2
4r^2 + 4r^2 +40r + 100 = 2500
8r^2 + 40r - 2400 = 0
r^2 + 5r - 300 = 0
(r+20)(r-15) = 0
r=-20 or r=15
Since this is a rate question, you throw out the negative rate, leaving the answer of r=15.

So, the first train was going 20 km per hour and the second train was going 15 km per hour.

Answer 4

v1=2*(x+5) (2*(x+5))^2+(2x)^2=50^2 then you will find the x
v2=2*(x)
answer is v1=x+5
v2=x
second problem:
110x+120y=46
110x+120y=115x+115y
as a resut x=y
110x+120x=46
x=y=0.2 so 110x=22 120y=24
answer is that:22 and 24/i have to go outside now if you mail me im gonna help you for last question see you

Answer 5

I have NO idea what you are asking. Math is my lowest subject. Reading what you wrote, I feel like I was reading something written in a strange language.
I suggest you try to answer them yourself. That is how you learn. Then, when you come up with the answers, ask your question again, with YOUR answers and see if others agree with what you came up with.

Answer 6

I dont tell the answers but I will give the concept to you.

If you want the concept mail me.

Answer 7

I'm only 12 so you're out of luck.

Answer 8

do ur own math work gal! im on break! summer break, i'll have enough of that when i go back!

Answer 9

...