3,50 degrees hope This heal
Hello :
the discriminat of each quadratic equation : ax²+bx+c=0 ....(a <span>≠ 0) is :
</span><span>Δ = b² -4ac
1 ) </span>Δ > 0 the equation has two reals solutions : x = (-b±√Δ)/2a
2 ) Δ = 0 : one solution : x = -b/2a
3 ) Δ <span>< 0 : no reals solutions</span>
Answer:
C. (-3,11)
Step-by-step explanation:
Tp is (-3,6) implies the quadratic could have been
f(x) = (x+3)²+6
(2/3)f(x) = (2/3)[(x+3)²+6]
= (2/3)(x+3)²+4
(2/3)f(x)+3 = (2/3)(x+3)²+4+3
= (2/3)(x+3)²+7
Tp at (-3,7)
Alternately,
No change in domain so x remains-3
(2/3)f(x) changes y from 6 to 4 (6×2/3)
+3 increases the y by 3
i.e 4+3 = 7
So, (-3,7)
Answer:
Perpendicular(p)=21
base(b)=20
hypotenuse(h)=29
Step-by-step explanation:
Now by using the formula:
cos A=b/h
i.e 20 by 29
The correct expression is A.
