Distance<-------------300--------------><-------------------x----------------->
⊕---------------------------------⊕--------------------------------------⊕ Express Freight Overtaking point
Speed 55 speed 30
Let x be the distance travelled by Freight until the point of overtaking
Time = Distance/Speed, then:
Time (freight) = Distance (fright)/Speed (freight) And
Time (Express) = Distance (Express)/Speed (Express).
Now plug in the relative number:
Time (freight) = x/30 and Time (express) = (x+300)/55. But note that both times are equal, then:
x/30 = (x+300)/55 OR (x÷30 = (x+300)÷55) [Answer C]
Now let's find the distance x. Cross multiplication;
55x= 30(300+x) →55x = 9,000 + 30x
55x - 30x = 9,000 → 25x = 9,000 and x = 9,000/25 = 360 miles
Time needed for Freight to travel 360 m = 360/30= 12 Hours
Time needed for Express to travel (300+360) m = 660/55= 12 Hours
As you see, it's the same time of 12 hours
Don't forget that Distance = Speed x time or time = Distance/Speed, etc.
Answer:
D
Step-by-step explanation:
From any point (x, y) on the parabola the focus and directrix are equidistant
Using the distance formula
= | y + 1 |
Squaring both sides
(x + 5)² + (y - 5)² = (y + 1)^2 , that is
(y + 1)² = (x + 5)² + (y - 5)² ← subtract (y - 5)² from both sides
(y + 1)² - (y - 5)² = (x + 5)² ← expand left side and simplify
y² + 2y + 1 - y² + 10y - 25 = (x + 5)²
12y - 24 = (x + 5)² ← factor left side
12(y - 2) = (x + 5)² ← divide both sides by 12
y - 2 = 
(x + 5)² ← add 2 to both sides
y = (x + 5)² + 2
or
f(x) = (x + 5)² + 2 → D
Answer:
b = (d-c)/a
Step-by-step explanation:
ab + c = d
We want to solve for b, so we need to get b alone
Subtract c from each side
ab+c-c = d-c
ab = (d-c)
Divide each side by a
ab/a = (d-c)/a
b = (d-c)/a
Answer:
40197 these many 5s make up 200985 so 4 are left to make 200989
Step-by-step explanation:
This is the concept of algebra, to solve the expression we proceed as follows;
cos 2x-cosx=0
cos 2x=cosx
but:
cos 2x+1=2(cos^2x)
thereore;
from:
cos 2x=cos x
adding 1 on both sides we get:
cos 2x+1=cos x+1
2(cos^2x)=cosx+1
suppose;
cos x=a
thus;
2a^2=a+1
a^2-1/2a-1/2=0
solving the above quadratic we get:
a=-0.5 and a=1
when a=-0.5
cosx=-0.5
x=120=2/3π
when x=1
cos x=1
x=0
the answer is:
x=0 or x=2/3π
