For the answer to the question above, first find out the gradient.
<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>
<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>
<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>
<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
<span>y-53 = 3x - 9 </span>
<span>y = 3x - 9 + 53 </span>
<span>y = 3x + 44 </span>
<span>y is the velocity of the car, x is the time.
</span>I hope this helps.
We have that the best estimate for the mass of a typical lineman on a high school football team is

From the options
1. 100,000 g
2. 100,000 kg
3. 100,000 mg
Generally
A Typical lineman cannot be as small as

A Typical linesman cannot be as massive as
as that is unrealistic
With the Average Mass of A Typical linesman can be

In conclusion
The best estimate for the mass of a typical lineman on a high school football team is

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star wars 1313 is a cancelled video game ( action / adventure ) that was under development by lucas arts.
Answer:
Running is probably the simplest individual sport for promoting both health-related and skill-related fitness. Fencing is often referred to as "mental chess" because it requires dexterity, endurance, and discipline. Individual sports require the coach to make all of the decisions for the athlete.
Answer:
a.) magnitude __49.7__ unit(s)
b.) direction __123.6°_ counterclockwise from the +x axis
Explanation:
Let Vector is v
x-component of Vector v = x = -27.5 units (minus sign indicate that x-component is along the minus x-axis )
y-component of Vector v = y = 41.4 units
Magnitude of v = ?
Direction of v = ?
To find the magnitude of the vector
v =
v = 
v = 49.7 units
To find direction
θ = tan⁻¹(y/x)
θ = tan⁻¹(41.4/-27.5)
θ = -56.4°
This Angle is in the clockwise direction with respect to -x axis.
We need to find Angle counterclockwise from the +x axis.
So,
θ = 180° - 56.4°
θ = 123.6°
The given vector is in 2nd quadrant