Please help with my geometry!!!!

Find expressions for the partial derivatives of the following functions:

nlexa [21]

Answer:

Step-by-step explanation: partial derivative is the differentiation of one variable e.g. X while leaving the values of the other variable e.g. Y

These four questions A, B, C and D have different functions separated by commas. I will not assume the commas to be something else like a plus sign.

A. f(x) = g'(x).k(y) , g'(x) + h(y)

f(y) = k'(y).g(x) , g(x) + h'(y)

B. f(x) = g'x (x+y)

f(y) = g'y (x+y) , h'y (y+z)

f(z) = h'z (y+z)

C. f(x) = f'x (xy) , f'x (zx)

f(y) = f'y (xy) , f'y (yz)

f(z) = f'z (yz) , f'z (zx)

D. f(x) = f'x (x) , g'(x) , h'x (x,y)

f(y) = h'x (x,y)

These are the partial derivative expressions for each variable in each function. You will need to pay a lot of attention to understand:

* while differentiating X alone, functions in Y which are separated by commas from the functions in X, are ignored totally because they are different questions

* In functions where X added to Y is in a bracket e.g. (x+y), to find the derivative of X, Y isn't thrown away because they are joined (by a plus sign) the derivative of X alone in this case would be f'x (x+y)

* f(x), just like g(x), simply means/represents a function in X hence f'(x) means the differentiation of all X-terms in that function

6 0

3 years ago

Find the area of the right triangle △DEF with the points D (0, 0), E (1, 1), and F.

Tems11 [23]

Answer:

The area of the triangle is $\sqrt{3}$

Step-by-step explanation:

Given:

Coordinates D (0, 0), E (1, 1)

Angle ∠DEF = 60°

△DEF is a Right triangle

To Find:

The area of the triangle

Solution:

The area of the triangle is = $\frac{1}{2}(base \times height)$

Here the base is Distance between D and E

calculation the distance using the distance formula, we get

DE = $\sqrt{(0-1)^2 + (0-1)^2}$

DE = $\sqrt{(-1) ^2 + (-1)^2$

DE = $\sqrt{1+1}$

DE = $\sqrt{2}$

Base = $\sqrt{2}$

Height is DF

DF = $tan(60^{\circ}) \times DE$

DF = $\sqrt{3} \times DE$

DF = $\sqrt{3} \times\sqrt{2}$

Now, the area of the triangle is

= $\frac{1}{2}({\sqrt{2})(\sqrt{3} \times \sqrt{2})$

= $\frac{1}{2}({\sqrt{2})(\sqrt{3} \sqrt{2})$

= $\frac{1}{2}(2\sqrt{3} )$

= $\sqrt{3}$

6 0

3 years ago

If the line segment at right is subdivided

Stells [14]

Answer:

50

Step-by-step explanation:

75 dived into 3 parts can also be written as 75/3. 75/3 is 25 making each part worth 25. 2 segments are included in the marked portion so its 50 or (25 + 25)

8 0

2 years ago

A rectangle has an area of 2x2 - 19x +45 ft? What are the dimensions of the rectangle? (hint: factor)

makkiz [27]

I think the answer is (2x - 9)(x - 5)

7 0

2 years ago

The owner of Darkest Tans Unlimited in a local mall is forecasting the demand for November the one new tanning booth based on th

lorasvet [3.4K]

Answer:

The forecast for November is 235 if August's forecast was 145.

Step-by-step explanation:

The formula for calculating forecast using exponential smoothing is:

$F_{t} = F_{t-1} + \alpha (A_{t-1} - F_{t-1} )$

Where Ft = New month forecast

Ft-1 = Previous month forecast

At-1 = Previous month actual value

α = smoothing constant

We are given F₈ = 145 (forecast for August), A₈ = 200 (Actual Value for August), α = 2, and we need to compute the forecast for November. So, We will first calculate the forecast for September then October and then November, step-by-step.

So, forecast for September is:

F₉ = F₈ + α (A₈ - F₈)

= 145 + 2*(200-145)

= 145 + 2*55

F₉ = 255

Then, forecast for October is:

F₁₀ = F₉ + α (A₉ - F₉)

= 255 + 2*(220-255)

= 255 + 2*(-35)

F₁₀ = 185

The forecast for November is:

F₁₁ = F₁₀ + α (A₁₀ - F₁₀)

= 185 + 2*(210 - 185)

F₁₁ = 235

8 0

3 years ago