Answer:
k would be 55
Step-by-step explanation:
35+90=125 and assuming that you're looking for a supplementary angle, the total must equal 180 so subtract 180-125 and you get k=55
93 is 60 percent from the original price.
The discount reduces the original price by 40 percent.
Answer:
Step-by-step explanation:
In order to find the horizontal distance the ball travels, we need to know first how long it took to hit the ground. We will find that time in the y-dimension, and then use that time in the x-dimension, which is the dimension in question when we talk about horizontal distance. Here's what we know in the y-dimension:
a = -32 ft/s/s
v₀ = 0 (since the ball is being thrown straight out the window, the angle is 0 degrees, which translates to no upwards velocity at all)
Δx = -15 feet (negative because the ball lands 15 feet below the point from which it drops)
t = ?? sec.
The equation we will use is the one for displacement:
Δx = 
and filling in:
which simplifies down to
so
so
t = .968 sec (That is not the correct number of sig fig's but if I use the correct number, the answer doesn't come out to be one of the choices given. So I deviate from the rules a bit here out of necessity.)
Now we use that time in the x-dimension. Here's what we know in that dimension specifically:
a = 0 (acceleration in this dimension is always 0)
v₀ = 80 ft/sec
t = .968 sec
Δx = ?? feet
We use the equation for displacement again, and filling in what we know in this dimension:
Δx = 
and of course the portion of that after the plus sign goes to 0, leaving us with simply:
Δx = (80)(.968)
Δx = 77.46 feet
Explanation:
We usually use graphs to solve two linear equations in two unknowns.
The basic idea is that a graph of an equation is the pictorial representation of all of the points that satisfy the equation. So, where the graph of one equation crosses the graph of another, the point where they cross will satisfy both equations.
Finding a solution means finding values of the variables that satisfy all of the equations. Hence, the point of intersection is the solution of the equations.
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To solve linear equations by graphing, graph each of the equations. Then find the coordinates of the point where the lines intersect. Those coordinates are the solution to the equations.
If the solution is not at a grid point on the graph, determining its exact value may not be easy. This can often be aided by a graphing calculator, which can often tell you the point of intersection to calculator accuracy.
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If the lines don't intersect, there are no solutions. If they are the same line (intersect everywhere), then there are an infinite number of solutions.
