I got you. The answer would be A (Certificate)
Answer:
7.5
Step-by-step explanation:
Let
x-------------> t<span>he width of each postcard
y-------------> the </span>the length of each postcard
we know that
A=12 in²
A=x*y-------> 12=x*y-------> equation 1
y=x+4-----> equation 2
<span>I substitute 2 in 1
</span>12=x*y----------> 12=x*[x+4]------> 12=x²+4x------> x²+4x-12=0
x²+4x-12=0
using a graph tool----> to resolve the second order equation
see the attached figure
the solution is
x=-6----> is not solution
x=2 in
y=x+4----> y=2+4-----> y=6 in
the answer isthe length of each postcard is 6 inthe width of each postcard is 2 in
Answer: The speed of car B is 6.44 m/s
Step-by-step explanation:
Let's define:
Sa = Speed of car A.
Sb = Speed of car B.
I assume that at t = 0s, the position of car A is 0m (then at t = 0s, the position of car B is 174m)
Then we can write the positions of each car as:
Pa(t) = Sa*t
Pb(t) = Sb*t + 174m.
We also know that Sa = 2*Sb.
And we know that at t= 27s, car A passes car B.
Then at t = 27s, the positions of both cars must be the same:
Pa(27s) = Pb(27s)
Sa*27s = Sb*27s + 174m.
Now we can replace Sa by 2*Sb (from the above equation)
2*Sb*27s = Sb*27s + 174m
Now we can solve this for Sb.
2*Sb*27s - Sb*27s = 174m
Sb*27s = 174m
Sb = 174m/27s = 6.44m/s.
The speed of car B is 6.44 m/s
Answer:
Step-by-step explanation:
〖{[√(1&20×5/9)×6^23×5]^230+5} 〗^25/√(5^25×25^5 )
I don't know what the & in the √(1&20×5/9) term means