Answer:
Step-by-step explanation:
Notice how it passed through at exactly (3,2) and (6,4).
By taking the formula for slope:
Substituting this into the slope-intercept form
y = mx + b
where m is the slope of the line and b is the y-intercept.
Since we were not given the y-intercept and the line started or passes through (0,0), we can safely assume that b = 0.
Therefore,
Are you sure you wrote that equation right
<h3>
Answer: 65</h3>
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Explanation:
We'll need to compute the difference quotient. In this case, we need to find what is equal to. It's called a difference quotient because there's a subtraction in the numerator (aka "difference") and we're dividing to form the quotient.
The idea is that as h approaches 0, then that expression I wrote will approach the derivative we're after. Keep in mind that h will technically never get to 0 itself. It only gets closer and closer.
Anyways, let's compute first
Then we'll subtract off g(t)
A very important thing to notice: the terms that don't have any 'h's in them have been canceled out (eg: 5t^2 combined with -5t^2 added to 0). Why is this important? It's because we need to factor 'h' out and we'll have a pair of 'h's cancel like so
The left hand side cannot have h = 0, or else we have a division by zero error. But if we approached 0 (not actually getting there), then the expression 5h+10t+5 will approach 5(0)+10t+5 = 10t+5
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In short: The derivative of is
In terms of symbols,
Later on in calculus, you'll learn a shortcut so you won't have to compute the difference quotient every time you need a derivative. Refer to the power rule for more information.
After we find the derivative, it's as straight forward as plugging in t = 6 to compute g ' (6)
Side note: This tells us that the slope of the tangent line is m = 65 when t = 6. In other words, this line is tangent to g(t) when t = 6, and this particular tangent line has slope m = 65.
Answer: Boat A should travel 152.8° to reach B
Step-by-step explanation:
The diagram illustrating the scenario is shown in the attached photo. Triangle ABP is formed. A represents the position of boat A. B represents the position of boat B. P represents the position of the port.
We would determine AB by applying the law of cosines
AB² = AP² + BP² - 2AP×BPCosP
AB² = 20² + 25² - 2 × 20 × 25 × Cos45
AB² = 1025 - 707.10678 = 317.89322
AB = √317.89322 = 17.83
We would determine the bearing of B from A by finding angle A. We would apply the sine rule.
AB/SinP = AP/Sin A
17.83/Sin45 = 20/SinA
Cross multiplying, it becomes
17.83 × SinA = 20Sin45 = 14.14
SinA = 14.14/17.83 = 0.79
A = Sin^-1(0.79) = 52.2°
The total angle at A is 65 + 52.2 = 117.2°
The angle formed outside the third quadrant is 117.2 - 90 = 27.2°
Therefore, bearing B from A is
180 - 27/2 = 152.8°