Answer: y=Mx+b
so y=-2/3x+5
Step-by-step explanation:
slope you go down 2 right 3, y int is (0,5)
Interesting problem ...
The key is to realize that the wires have some distance to the ground, that does not change.
The pole does change. But the vertical height of the pole plus the distance from the pole to the wires is the distance ground to the wires all the time. In other words, for any angle one has:
D = L * sin(alpha) + d, where D is the distance wires-ground, L is the length of the pole, alpha is the angle, and 'd' is the distance from the top of the (inclined) pole to the wires:
L*sin(40) + 8 = L*sin(60) + 2, so one can get the length of the pole:
L = (8-2)/(sin(60) - sin(40)) = 6/0.2232 = 26.88 ft (be careful to have the calculator in degrees not rad)
So the pole is 26.88 ft long!
If the wires are higher than 26.88 ft, no problem. if they are below, the concerns are justified and it won't pass!
Your statement does not mention the distance between the wires and the ground. Do you have it?
So, you have to have your equation in terms of dimes. What you said about what he has: four more nickels than dimes in his pocket. Will help with our equation.
We know what a nickel and dime is worth. A nickel is .05 and a dime is .1
We don't know how many dimes are in his pocket, since we're trying to solve it.
.1x+(4+x).05=1.25;
In the parenthesis, it shows how there are four more nickels. Let's solve it now.
.1x + (4+x).05=1.25
.1x + .2+ .05x = 1.25; Let's add like terms.
.15x + .2 = 1.25; Subtract .2 from both sides.
.15x= 1.05
1.05÷.15= 7 =x
There are 7 dimes in his pocket, let's check our answer.
We now know there are 11 nickels, since there are four more nickels than dimes.
11(.05) +.1(7) = 1.25
.55+ .7 = 1.25
Now that we've tried it, we know there are 7 dimes in his pocket.
Tell me if this helps!
The formula of the line is
y=mx + b
The likes goes through the point (0,0)
that means that y-int = 0, b=0
y=mx
So we need to find only slope.
The answer is choice A.
We're told that the left and right walls of the cube (LMN and PQR) are parallel planes. Any line contained in one of those planes will not meet another line contained in another plane. With choice A, it's possible to have the front and back walls be non-parallel and still meet the initial conditions. If this is the case, then OS won't be paralle to NR. Similarly, LP won't be parallel to MQ.
