Answer:
for quadratic equation 1
y^2-10y+21=0
Step-by-step explanation:
y^2-7y-3y+21=0
y(y-7)-3(y-7)=0
(y-7)(y-3)=0
y-7=0,y-3=0
y=7,y=3
y=(3,7)
for quadratic equation 2
16p^2-8p+1=0
16p^2-4p-4p+1=0
4p(4p-1)-1(4p-1)=0
(4p-1)twice=0
4p-1=0,4p-1=0
4p=1
p=1/4 twice
for quadratic equation 3
x^2-400=0
x^2=400
x=√400
x=20
for quadratic equation 4
-16m^2-8m-1=0
multiply the equation by -
16m^2+8m+1=0
16m^2+4m+4m+1=0
4m(4m+1)1(4m+1)=0
4m+1=0 twice
m=-1/4 twice
for quadratic equation 5
-3n^2+75=0
divide both side by -3
-3n^2/-3=-75/-3
n^2=25
n=√25
n=5
Answer:
The answer is 4.
Step-by-step explanation:
1. find the factors of 16(1,2,4,4,8,16) and of 28(1,2,4,7,14,28)
2. find the highest factor of the two Factored #.(4)
Answer:
1 hour
Step-by-step explanation:
I find these easiest to work by considering the initial difference in distance and the speed at with that gap is closing.
__
The gap is 15 miles, the distance the first ship is from harbor when the second ship starts.
The rate of closure is the difference in the speeds of the two ships:
60 mph -45 mph = 15 mph
Then the closure time is ...
time = distance/speed
time = (15 mi)/(15 mi/h) = 1 h
It will take the second ship 1 hour to catch up to the first ship.
