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3 years ago

D= kA [T2 - T1 / L] Solve for T1

1 answer:

5 0

$d=\dfrac{k_A(T_2-T_1)}{T}\ \ \ \ |multiply\ both\ sides\ by\ T\\\\k_A(T_2-T_1)=dT\ \ \ \ |divide\ both\ sides\ by\ k_A\\\\T_2-T_1=\dfrac{dT}{k_A}\ \ \ \ |subtract\ T_2\ from\ both\ sides\\\\-T_1=\dfrac{dT}{k_A}-T_2\ \ \ \ \ |change\ signs\\\\\boxed{T_1=T_2-\frac{dT}{k_A}}$

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Which expression is equal to 2x7x6?

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2 years ago

A birthday celebration meal is $ 47.40 including tax, but not the tip. Find the total cost if a 15% tip is added to the cost of

Afina-wow [57]

Answer:

$54.51

Step-by-step explanation:

an easier way to do this is simply by multiplying 47.40x115%, which is 54.51.

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3 years ago

Factor the expressions .<br> 36 + 8z

Sophie [7]

Answer:

4 × ( 9 + 2z)

Step-by-step explanation:

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we can also write it like this

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3 years ago

maya has 15 stickers . she places them on a page in rows of 5 . there are 2 flowers and 3 hearts in each row . how many hearts a

MrMuchimi

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3 years ago

Read 2 more answers

Find the slope of the tangent line to the curve f(x)=e^(x) at (0.4,1.49)

almond37 [142]

Answer:

1.49

Step-by-step explanation:

In order to find the slope of the tangent line to a given equation, and in a given point, we need to:

1. Find the first derivative of the given function.

2. Evaluate the first derivative function in the given point.

1. Let's find the first derivative of the given function:

The original function is $f(x)=e^{x}$

But remeber that the derivative of $e^{x}$ is $e^{x}$

so, $f'(x)=e^{x}$

2. Let's evaluate the first derivative function in the given point

The given point is (0.4,1.49) so:

$f'(x)=e^{x}$

$f'(0.4)=e^{0.4}$

$f'(x)=1.49$

Notice that the calculated slope of the tangent line is equal to the y-coordinate of the given point because f'(x)=f(x). In conclusion, the slope of the tangent line is equal to 1.49.

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3 years ago