1. 3x - 2y = 0 => y = 3/2x
4x + 2y = 14<=> 4x + 2*(3/2x) = 14<=> 7x = 14<=> x = 2 & y = 3
2.3p + q = 7=> q = 7 - 3p
2p - 2q = -6<=> 2p - 2*(7-3p) = -6<=> 2p - 14 + 6p = -6<=> 8p = -6 + 14 = 8<=> p = 1 & q = 4
3.3x - 2y = 1=> x = (1+2y)/3
8x + 3y = 2<=> 8*(1+2y)/3 + 3y = 2<=> 8*(1+2y)/3 - 2 = -3y<=> 3*(8*(1+2y)/3 - 2) = -3*(3y)<=> 8*(1+2y) - 6 = -9y<=> 8 + 16y - 6 = -9y<=> 2 = -25y<=> y = -2/25 & x = 7/25
I believe so. one example i can think of is a binomial(idk if it counts as a polynomial tho bc some schools learn different things)such as x+5 multiplied by another binomial such as x+4. (x+5)(x+4) = x^2+9x+20 which is a trinomial.
Correct is:
B ) The slope of a line which moves up from left to right is positive.
C ) Some lines have no slope.
Answer:
The distance between the airplane and the airport is:
Step-by-step explanation:
We can use the Pythagoras theorem to find the distance from the airplane to the airport.
The equation is:
Where:
- <u>a represents the elevations of the airplane (15000 feet)</u>
- <u>b is the distance from the ground directly below the airplane to the airport (6 miles)</u>
- <u>c is the distance between the airplane and the airport.</u>
We need first to convert 15000 feet in miles.
Therefore, c will be:
I hope it helps you!