MATHEMATICS COLLEGE

John's electricity bill costs $20.90 per month plus $1.23 per kilowatt hour. How many kilowatt hours can he use and keep his monthly cost no more than $109

Answers

Answer 1

Answer:

71.63 kilowatts hours

Step-by-step explanation:

We are told in the question:

John's electricity bill costs $20.90 per month plus $1.23 per kilowatt hour.

We are to find, How many kilowatt hours can he use and keep his monthly cost no more than $109.

Step 1

$109 - $20.90

= $88.1

$88.1 is the amount left to spend on killowatts of electricity per hour after removing the normal monthly electricity bill in a month

Step 2

$1.23 = 1kilowatts per hour

$88.1 = y kilowatts per hour

Cross Multiply

= $1.23 × y kilowatts per hour = $88.1 × 1 kilowatts per hour

y kilowatts per hour = $88.1 × 1 kilowatts per hour/ $1.23

= 71.62601626kilowatts hour.

Approximately = 71.63 kilowatts hour

Therefore, John can use 71.63 kilowatts per hour and keep his monthly cost no more than $109

Answer 2

Answer:

Answer:

kilowatt hours can he use and keep his monthly cost no more than $109

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Compute the area of the region D bounded by xy=1, xy=16, xy2=1, xy2=36 in the first quadrant of the xy-plane. Using the non-linear change of variables u=xy and v=xy2, find x and y as functions of u and v.x=x(u,v)= ?

y=y(u,v)=?

Find the determinant of the Jacobian for this change of variables.

∣∣∣∂(x,y)/∂(u,v)∣∣∣=det=?

Using the change of variables, set up a double integral for calculating the area of the region D.

∫∫Ddxdy=?

Evaluate the double integral and compute the area of the region D.

Area =

Answers

Answer:

53.7528

Step-by-step explanation:

Notice that when

$xy = 1 ,\,\,\, xy = 16 , \,\,\, xy^2 = 1 \,\,\,, xy^2 = 36 \n\n$

If you set

$u = xy , v = xy^2$

as they suggest, then

$y = (v)/(u)} \,\,\,\, \text{and} \,\,\,\, \n\n{\displaystyle x = (u)/(y) = (u)/(v/u) = (u^2)/(v)$

Then

${\diplaystyle (\partial(x,y))/(\partial(u,v))} =\det \begin{pmatrix} 2u/v && -u^2/v^2 \n -v/u^2 && 1/u \end{pmatrix} = (1)/(v) }$

Therefore

$\iint\limits_(D) dx\,dy = \int\limits_(1)^(36)\int\limits_(1)^(16) (1)/(v) \, du \, dv = 15 \ln(36) = 53.7528$

A Jacobian matrix is formed by the first partial derivatives of a multivariate function that utilizes a training algorithm, and further calculation as follows:

Jacobian:

To evaluate the integral, cover the bounds, the integrand, and the differential area dA.

Transform the four equations in terms of u and v, notice that $u= xy \ \ and \ \ xy = 1, xy = 16$

implies that $1\leq u \leq 16.$

Similarly, $v= xy^2\ \ and\ \ xy^2= 1 , xy^2= 25$ implies that $1 \leq v \leq 25$

so write this integration region as $S= {(u,v) |1 \leq u \leq 18, 1 \leq v \leq 25}.$

Translate the equations from uv - plane to xy- plane. It is obtained by solving,

$u= xy, y= xy^2 \n\n\left.\begin{matrix}u=xy & \n v=xy^2& \end{matrix}\right\} \to \left.\begin{matrix}u^2=x^2y^2 & \n v=xy^2& \end{matrix}\right\} \n\n\to x=(u^2)/(v), y=(v)/(u)$

Convert dA part of the integral , using is $dA= |(\partial (x,y))/(\partial(u,v))| dudv.$

That is, $dA= \begin{vmatrix}(\partial x)/(\partial u) & (\partial x)/(\partial v)\n (\partial y)/(\partial u) & (\partial y)/(\partial v) \end{vmatrix} \ du dv \n\n$

Sampule the partial derivatives to find the Jacobian.

$dA=\begin{vmatrix}(2u)/(v) &-(u^2)/(v^2) \n -(v)/(u^2) &(1)/(u) \end{vmatrix} \ dudv\n\n=[((2u)/(v)) ((1)/(u)) -(- (u^2)/(v^2))(-(v)/(u^2))]\ du dv\n\n=((2)/(v)- (1)/(v)) \ dudv\n\n=(1)/(v)\ du dv\n\n$

The Jacobian the transformation is $dA= (1)/(v)dudv$

The region is $S={(u,v) |1\leq u \leq 16, 1\leq v\leq 25}.$

Rewrite the integral, using the transformation: $S,\ x=(u^2)/(v) =, y=(v)/(u) \ \ and\ \ dA=(1)/(v) dudv\n\n\int\int_R 1dA =\int \int_S (1)/(v)\ dudv= \int^(25)_(1) \int^(16)_(1) \ (1)/(v) \ dudv\n\n$

Evaluate the inner integral with respect to u.

$\to \int\int_R 1dA = \int^(25)_(1) \int^(16)_(1) \ (1)/(v) \ dudv\n\n$

by solving the value we get

$= 30 \ ln (5) \approx 48.28$

Find out more about the Jacobians here:

brainly.com/question/9381576

Mang Jose plans to fence his rectangular lot before he will plant mushrooms for his mushrooms production business. The perimeter of the lot is 40 meters and the area is 96 square meters.Using the concept of the sum and product of roots of a quadratic equation, how would you determine the length and the width of the rectangular lot? Provide a quadratic equation representing this scenario.

Someone help me

Answers

Answer: The rectangular lot is 12x8 meters

Step-by-step explanation:Perimeter of a geometric figure is the sum of all its sides.

A rectangle is a quadrilateral that has opposite sides parallel and equal, which means, and suppose l is length and w is width:

P = 2l + 2w

The perimeter of the lot is 40m, thus:

2l + 2w = 40

Area of a rectangle is calculated as:

A = length x width

The lot has area of 96, thus:

lw = 96

Solving the system of equations:

2l + 2w = 40 (1)

lw = 96 (2)

Isolate l from (1):

2l = 40 - 2w

l = 20 - w (3)

Substitute (3) in (2):

w(20-w) = 96

$-w^(2)+20w=96$

$-w^(2)+20w-96=0$

There are many methods to determine the roots of a quadratic equation. One of them is using the sum and product of those roots.

Sum of the roots is given by:

$sum = (-b)/(a)$

$sum=(-20)/(-1)$

sum = 20

Product of the roots is:

$prod=(c)/(a)$

$prod=(-96)/(-1)$

prod = 96

The roots of the quadratic equation are numbers which the sum results in 20 and product is 96:

w₁ = 12

w₂ = 8

If we substitute w to find l, the numbers will be l₁ = 8 and l₂ = 12.

Since length is bigger than width, the rectangular lot Mang Jose has to plant mushrooms measures 12m in length and 8m in width

A stock of food is enough to fees 50 persons for 14 days. How many days will the foos last if 20 persons will be added?

Answers

Answer:

solution

Step-by-step explanation:8 days

Find the real value of x,y if

(1/(x^2+y^2))+(1/(x+yi))=1

Answers

It sounds like $x,y$ are supposed to be real numbers. If so, then we can do the following.

$\frac1{x^2+y^2}+\frac1{x+yi}=1$

Multiply the second term's numerator and denominator by the conjugate of the denominator:

$\frac1{x^2+y^2}+(x-yi)/((x+yi)(x-yi))=1$

$\frac1{x^2+y^2}+(x-yi)/(x^2+y^2)=1$

$(x+1-yi)/(x^2+y^2)=1$

Since the left hand side is equal to 1, this means it has no imaginary part, so that $y=0$ . Then the real parts of both sides of the equation give us

$(x+1)/(x^2)=1\implies x+1=x^2\implies x^2-x-1=0\implies x=\frac{1\pm\sqrt5}2$

A decision maker is considering including two additional variables into a regression model that has as the dependent variable, Total Sales. The first additional variable is the region of the country (North, South, East, or West) in which the company is located. The second variable is the type of business (Manufacturing, Financial, Information Services, or Other). Given this, how many additional variables will be incorporated into the model?a. 9
b. 8
c. 6
d. 2