Answer:
C
Step-by-step explanation:
Given 2 secants intersect a circle from a point outside the circle, then
The product of the external part and the entire part of one secant is equal to the product of the external part and the entire part of the other secant, that is
x(x + 10 + x) = 6(6 + 10 + x)
x(2x + 10) = 6(16 + x) ← distribute parenthesis on both sides
2x² + 10x = 96 + 6x ← subtract 96 + 6x from both sides
2x² + 4x - 96 = 0 ← in standard form
Divide through by 2
x² + 2x - 48 = 0 ← factor the left side
(x + 8)(x - 6) = 0
Equate each factor to zero and solve for x
x + 8 = 0 ⇒ x = - 8
x - 6 = 0 ⇒ x = 6
However x > 0 ⇒ x = 6 → C
To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.
jk=ac and j+k=b
The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...
2k^2-5k-18=0
2k^2+4k-9k-18=0
2k(k+2)-9(k+2)=0
(2k-9)(k+2)=0
so k=-2 and 9/2
k=(-2, 4.5)
Answer:
Step-by-step explanation:
they want to know the high and low quiz scores... that's all
Answer:
C. 15
Step-by-step explanation:
-4/3x-6=-26
The first step is to add 6 to each side
-4/3x-6+6=-26+6
-4/3x =-20
Now multiply by -3/4 to isolate x.
-3/4 * -4/3x = -3/4 * -20
x = 60/4
x = 15
Answer: -3000
Step-by-step explanation:
That answer was on quizlet
