1. A quadratic equation has the following form: ax²+bx+c.
2. The leading coefficient is the number that is attached to the variable with the highest exponent. Then, the "a" is the leading coefficient of the quadratic equation.
3. The problem says that the leading coefficient is 1 (a=1) and the roots of the quadratic equation are 5 and -3. Then, you have:
(x-5)(x+3)=0
4. When you apply the distributive property, you obtain:
x²+3x-5x-15=0
x²-2x-15=0
5. Therefore, the answer is:
x²-2x-15=0
Answer:
495 meter²
Step-by-step explanation:
In the given kite QRST,
PQ = PS = 15 meter
QR = 17 meter
We have to find the area of the kite.
Since kite is in the form of a rhombus.
and rhombus is = 
(Diagonal QS) × (Diagonal RT)
In Δ QRP,
17² = 15² + RP²
= √64 = 8 meters.
So RT = RP + PT = 8 + 25 = 33 meter.
Now area of kite = (30) (33) = 495 meter²
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