Provide all of the names of the angle in the figure

The output of an economic system Q, subject to two inputs, such as labor L and capital K, is often modeled by the Cobb-Dougl

Anna35 [415]

Answer:

Step-by-step explanation:

Given

Economic system Q is given by

$Q=cL^aK^b$

also $a+b=1$

if $Q=12,200$

$a=\frac{1}{6}$

$b=\frac{5}{6}$

$c=42$

substitute these values

$12,200=42\times (L)^{\frac{1}{6}}K^{\frac{5}{6}}$

$(L)^{\frac{1}{6}}K^{\frac{5}{6}}=\frac{12,200}{42}$

$K^{\frac{5}{6}}=\frac{12,200}{42(L)^{\frac{1}{6}}}$

$K=(\frac{12,200}{42})^{\frac{6}{5}}\times \frac{1}{L^{5}}$

differentiate w.r.t to L to get $\frac{dK}{dL}$

$\frac{dK}{dL}=(\frac{12,200}{42})^{\frac{6}{5}}\times (-5)\times L^{-6}$

$\frac{dK}{dL}=-5(\frac{12,200}{42})^{\frac{6}{5}}\times \frac{1}{L^6}$

$\frac{dK}{dL}=-\frac{4515.466}{L^6}$

6 0

3 years ago

Jan'ai was asked to determine the minimum for a function with zeros located at –1 and 5, which also has a y-intercept of (0, –25

sergey [27]

The first error in Jan'ai's work in determining the considered function is given by: Option D: She incorrectly determined the x-coordinate of the vertex.

<h3>What are the coordinates of vertex for a quadratic function?</h3>

For a quadratic function of the form $y = ax^2 + bx + c$ , its vertex form is obtained as:

$y = ax^2 + bx + c\\y =a(x^2 + bx/a) + c\\y = a(x^2 + 2(b/2a)x + (b/2a)^2 -(b/2a)^2 )+ c\\y = a(x^2 + 2(b/2a)x + (b/2a)^2) -a \times (b/2a)^2 + c\\\\y = a(x+b/2a)^2 - a \times (b/2a)^2 + c$

For the form $y = a(x-h)^2 + k$ , the vertex has coordinates (h, k)

Thus, for the obtained equation $y = a(x+b/2a)^2 - a \times (b/2a)^2 + c$ , we get the coordinates of vertex as:

$h = -b/2a$ , $k = c - a\times(b/2a)^2$

Thus, the coordinates of vertex of $y = ax^2 + bx + c$ is:

$(h,k) = (-b/2a, c - a \times (b/2a)^2 )$

The missing steps of work of Jan'ai are:

Begin to write a function in factored form. $f(x) = a(x+1)(x-5)$
Substitute x = 0, y = f(x) = -25 to determine a. $-25 = a(0+1)(0-5)$
Simplify and solve to find a. $a=5$
Rewrite the function. $f(x) = 5(x+1)(x-5)$
Rewrite in standard form. $f(x) = 5x^2-20x-25$
Find the x-coordinate of the vertex. $x = -20/2(5) = -20/10; x = -2$
Find the y-coordinate of the vertex.

$y = 5x^2-20x-25$

$y = 5(-2)^2-20(-2)-25$

$y = 35$

so (-2,35) is the coordinate of the vertex, which denotes the minimum.

So, as we see, in the 5th step, Jan'ai had the quadratic function $f(x) = 5x^2-20x-25$ ,

Comparing this to $f(x) = ax^2 + bx + c$ , we get a = 5, b = -20, c = -25

The vertex's x-coordinate will be on -b/2a = -(-20)/ 2(5) = 20/10 = 2

But Jan'ai didn't putted that negative sign before b. in the 6th step.

Thus, the first error in Jan'ai's work in determining the considered function is given by: Option D: She incorrectly determined the x-coordinate of the vertex.

Learn more about the vertex form of a quadratic equation here:

brainly.com/question/9912128

6 0

3 years ago

An angle measures 110.8° more than the measure of its supplementary angle. What is the measure of each angle?

stealth61 [152]

Answer:

34.6 and 145.4

Step-by-step explanation:

x+(x+110.8) = 180

2x= 180-110.8

2x= 69.2

x= 34.6

Measure of each angle:

x= 34.6

x+110.8 = 34.6 + 110.8 = 145.4

3 0

3 years ago

154 is the same as the product of 158 and n, decreased by 132

grandymaker [24]

Answer:

Equation: 154 = 158n - 132 n = 11/6

Step-by-step explanation:

Let's start by converting this into an equation. Some keywords to look for are "is" which means an equal sign (=), "product" which is multiplication (*), and "decreased" which means a minus sign (-):

So this is our equation: 154 = 158n - 132

Now we have to solve for n:

First, we add 132 to both sides to isolate the n more:

286 = 158n

Then to completely isolate the n, we divide both sides by 158

286

------ = n

156

<u>n = 11/6, final answer.</u>

8 0

2 years ago

Simplify -4|-5| - [2(-17)]

anastassius [24]

Answer:

The answer would be 14.

Step-by-step explanation:

You need to SIMPLIFY. Make sure that you are solving the part of the problem that has parentheses then you can move on from there.

(Hope this helps!)

4 0

3 years ago