Find two numbers whose product is 80 and whose sum is 18

Let X & Y be the two integers

X+Y=18      Y = 18-X

XY = 80

We can substitute 18-X for Y in the equation XY = 80

X(18-X) = 80

Distributing

18X - X2 = 80

Subtract 80 from both sides so that the equation is equal to 0

-X2 + 18X - 80 = 0

Multiply (or divide by -1 so that the leading coefficient is positive, not negative)

X2 - 18X + 80 = 0

We can factor this quadratic

(X-10)(X-8) = 0

Setting each factor equal to 0 we can solve for X.

X-10 = 0       X-8 = 0

X = 10          X = 8

The two integers are 8 and 10.

Step-by-step explanation:

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

answer: sine

step-by-step explanation: