Given the function f(x)=x^2+5x+5 , to get the zeros we solve using quadratic formula;
x=[-b(+or-) sqrt(b^2-4ac)]/(2a)
from our function;
a=1,b=5, c=5
thus,
x=[-5(+or-)sqrt(5^2-4*1*5)]/(2*1)
x=[-5(+or-)sqrt(25-20)]/2
x=[-5+/-sqrt(5)/2
The answer is option C
1) 13.02-6y=8y
+6y=6y
13.02=14y then divide both sides by 14. you get y= .93
2) 1/5x+5=19-1/2x
-5=-5
(1/5x=14-1/2x)5
x=70-2.5x
+2.5x=+2.5x
3.5x=70 divide both sides by 3.5 and you get x=20
3) 7.3t+22=2.1t-22.2
-2.1t =-2.1t
5.2t+22=22.2
-22=-22
5.2t=-44.2 divide both by 5.2 and t=-8.5
4) (1.4+2/5e=3/15e-.8)15
21+6e=3e-12
-3e=-3e
21+3e=-12
-21 =-21
3e=-33 divide both by 3 and e=-11
5) 5(x-4)=2(2x+5)
5x-20=4x+10
-4x -4x
x-20=10
+20=+20
x=30
6) -7(3+t)=4(2t+6)
-21-7t=8t+24
+7t=+7t
-21=15t+24
-24= -24
-45=15t divide both by 15 and t=-3
7) (3/4x+6=1/3x+9)4
3x+24=1.3x+36
-1.3x =-1.3x
1.7x+24=36
-24=-24
1.7x =12 divide both sides by 1.7 and x=7.06
They are 150 apart
all 3 are even
and they are all in the hundreds
Answer:
C
Step-by-step explanation:
