Answer:
1 and 2 SORRY IF I AM WRONG
EQ. 1: x + y = 8
EQ. 2: 4x-y = 7
Rewrite EQ. 1 as EQ. 3: x = 8-y
Replace x in EQ.2 with EQ. 3:
4(8-y) - y = 7
Use the distributive property:
32 - 4y - y = 7
Combine like terms:
32 - 5y = 7
Subtract 32 from each side:
-5y = -25
Divide both sides by -5
y = -25 / -5
y = 5
Now replace y with 5 in EQ. 3 to solve for X:
x = 8-5
x = 3
The point of intersection is X = 3, Y = 5, which is written as (3,5)
Answer:
It is B. hyperbola, 45 degrees.
SteIt is p-by-step explanation:
If we rotate the standard form x^2 - y^2 = 1 through 45 degrees we get xy = 1/2.
xy = -2.5 comes from x^2 - y^2 = -5 being rotated 45 degrees.
<span>N(t) = 16t ; Distance north of spot at time t for the liner.
W(t) = 14(t-1); Distance west of spot at time t for the tanker.
d(t) = sqrt(N(t)^2 + W(t)^2) ; Distance between both ships at time t.
Let's create a function to express the distance north of the spot that the luxury liner is at time t. We will use the value t as representing "the number of hours since 2 p.m." Since the liner was there at exactly 2 p.m. and is traveling 16 kph, the function is
N(t) = 16t
Now let's create the same function for how far west the tanker is from the spot. Since the tanker was there at 3 p.m. (t = 1 by the definition above), the function is slightly more complicated, and is
W(t) = 14(t-1)
The distance between the 2 ships is easy. Just use the pythagorean theorem. So
d(t) = sqrt(N(t)^2 + W(t)^2)
If you want the function for d() to be expanded, just substitute the other functions, so
d(t) = sqrt((16t)^2 + (14(t-1))^2)
d(t) = sqrt(256t^2 + (14t-14)^2)
d(t) = sqrt(256t^2 + (196t^2 - 392t + 196) )
d(t) = sqrt(452t^2 - 392t + 196)</span>
Answer: A=113.1cm
Step-by-step explanation:
