Answer: See pic above for answer. I got it from Photomath
Plz put brainliest
I am looking on the answers, and there is only one case, when a or b or c or d pass: 3|x-3| + 2 = 14. So I assume, that before two is plus. Then:
3|x-3|+2=14 |minus 2
3|x-3|=12 |divide 3
|x-3|=4
From absolute value definition you've got two ways:
x-3=4 or x-3=-4
x=7 or x=-1
And answer d) passes
Answer:
Your question is in what base please?
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
