One positive number is 5 less than twice a second number, and their product is 140. find the two numbers.

x, y - two positive numbers

one positive number is 5 less than twice a second number

(1)        x = 2y - 5

their product is 140

(2)    xy = 140

substitute from (1) to (2)

(2y - 5)y = 140      use distributive property

(2y)(y) - (5)(y) = 140

2y² - 5y = 140       subtract 140 from both sides

2y² - 5y - 140 = 0

use quadratic formula

a = 2, b = -5, c = -140

b² - 4ac = (-5)² - 4(2)(-140) = 25 + 1120 = 1145

substitute the value of y₁ to (1)

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

is it 35

step-by-step explanation: