Parallelogram Area = Height * Base
Base = Parallelogram Area / Height
Base = 36 / 3
Base = 12 centimeters
Answer:
Yes, by AA theory
Step-by-step explanation:
They are related right triangles.
<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
![g(t) = 5t^2+5t\\\\g(h+t) = 5(h+t)^2+5(h+t)\\\\g(h+t) = 5(h^2+2ht+t^2)+5(h+t)\\\\g(h+t) = 5h^2+10ht+5t^2+5h+5t\\\\](https://tex.z-dn.net/?f=g%28t%29%20%3D%205t%5E2%2B5t%5C%5C%5C%5Cg%28h%2Bt%29%20%3D%205%28h%2Bt%29%5E2%2B5%28h%2Bt%29%5C%5C%5C%5Cg%28h%2Bt%29%20%3D%205%28h%5E2%2B2ht%2Bt%5E2%29%2B5%28h%2Bt%29%5C%5C%5C%5Cg%28h%2Bt%29%20%3D%205h%5E2%2B10ht%2B5t%5E2%2B5h%2B5t%5C%5C%5C%5C)
Then we'll subtract off g(t)
![g(h+t)-g(t) = (5h^2+10ht+5t^2+5h+5t) - (5t^2+5t)\\\\g(h+t)-g(t) = 5h^2+10ht+5t^2+5h+5t - 5t^2-5t\\\\g(h+t)-g(t) = 5h^2+10ht+5h\\\\](https://tex.z-dn.net/?f=g%28h%2Bt%29-g%28t%29%20%3D%20%285h%5E2%2B10ht%2B5t%5E2%2B5h%2B5t%29%20-%20%285t%5E2%2B5t%29%5C%5C%5C%5Cg%28h%2Bt%29-g%28t%29%20%3D%205h%5E2%2B10ht%2B5t%5E2%2B5h%2B5t%20-%205t%5E2-5t%5C%5C%5C%5Cg%28h%2Bt%29-g%28t%29%20%3D%205h%5E2%2B10ht%2B5h%5C%5C%5C%5C)
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
![\frac{g(h+t)-g(t)}{h} = \frac{5h^2+10ht+5h}{h}\\\\\frac{g(h+t)-g(t)}{h} = \frac{h(5h+10t+5)}{h}\\\\\frac{g(h+t)-g(t)}{h} = 5h+10t+5\\\\](https://tex.z-dn.net/?f=%5Cfrac%7Bg%28h%2Bt%29-g%28t%29%7D%7Bh%7D%20%3D%20%5Cfrac%7B5h%5E2%2B10ht%2B5h%7D%7Bh%7D%5C%5C%5C%5C%5Cfrac%7Bg%28h%2Bt%29-g%28t%29%7D%7Bh%7D%20%3D%20%5Cfrac%7Bh%285h%2B10t%2B5%29%7D%7Bh%7D%5C%5C%5C%5C%5Cfrac%7Bg%28h%2Bt%29-g%28t%29%7D%7Bh%7D%20%3D%205h%2B10t%2B5%5C%5C%5C%5C)
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
---------------------
In short: The derivative of
is ![10t+5](https://tex.z-dn.net/?f=10t%2B5)
In terms of symbols, ![g ' (t) = 10t+5](https://tex.z-dn.net/?f=g%20%27%20%28t%29%20%3D%2010t%2B5)
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)
![g ' (t) = 10t+5\\\\g ' (6) = 10(6)+5\\\\g ' (6) = 60+5\\\\g ' (6) = 65\\\\](https://tex.z-dn.net/?f=g%20%27%20%28t%29%20%3D%2010t%2B5%5C%5C%5C%5Cg%20%27%20%286%29%20%3D%2010%286%29%2B5%5C%5C%5C%5Cg%20%27%20%286%29%20%3D%2060%2B5%5C%5C%5C%5Cg%20%27%20%286%29%20%3D%2065%5C%5C%5C%5C)
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.
Sorry can't see the pic really good
Answer:
9(3+c+4)=
27+9c+36=
63+9c=
Step-by-step explanation:
9(3+c+4)=
27+9c+36=
63+9c=