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3 years ago

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Suppose the cost of electricity is $0.03 for each kilowatt-hour. Carlos's house runs on 11.14 kilowatt-hours a day. Find the cos

t of electricity for Carlos's house in one day.

1 answer:

irga5000 [103]3 years ago

8 0

Answer: I think the answer is $33.42

Explanation: If it’s $0.03 for every hour and there is only 11.14 hours used a day just multiply 11.14 by .03 and you get .3342 then you move the decimal over 2 spaces to get 33.42

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George wants to invest $20,000 One investment pays interest at a rate of 18 2/3% while another pays interest at a rate of 0.1872

xz_007 [3.2K]

Answer:

Investment with interest rate 0.1872 is better

Step-by-step explanation:

Given as :

George investment principal = $20,000

The Rate of interest (R1) = 18 $\frac{2}{3}$ % = $\frac{56}{3}$ %

I.e R1 = 18.67 %

Again The The Rate of interest (R2) = 0.1872 = 18.72 %

Let the time period for both rate of interest = 1 years

Now from compound interest method

Amount = Principal × $(1 + \frac{Rate}{100})^{Time}$

Or, A 1 = $ 20 , 000 × $(1 + \frac{18.67}{100})^{1}$

Or, A 1 = $ 20 ,000 × 1.1867

∴ A 1 = $ 23,734

And A 2 = $ 20 , 000 × $(1 + \frac{18.72}{100})^{1}$

Or, A 2 = $ 20 ,000 × 1.1872

∴ A 2 = $ 23,744

Hence From the calculation of both amount it is clear that , investment with interest rate 0.1872 is better . Answer

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3 years ago

VERY DIFFICULT!! HELP NEEDED

Nataly [62]

Answer:

4 dogs....hope it helps u

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3 years ago

Plz help I’m getting timed

scoray [572]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

How do I get the answer y= 2x^2+9x-5

guapka [62]

Answer:

Quadratic Formula

so

x = -5

and

x = 0.5

Step-by-step explanation:

Whenever you see a problem in this form, which you will see a lot of, you can try to factor it or use the "least squares" method or what have you, but those won't always work, unfortunately.

Fortunately, the quadratic formula will never fail you with quadratic expressions.

This is the Quadratic Formula

$x = \frac{-b \frac{+}{-} \sqrt{b^2-4ac} }{2a}$

a is the the number on the variable with the exponent ^2

b is the number on the variable with no exponent

c is the third number

a and b cannot be equal to 0; c can be

Since we're looking for a number with an equation that has a square root in it, we're going to get two answers. These two answers come from the radical being separately added AND subtracted from the radical. It's basically two problems.

Plugging in our numbers to this equation gives us x values of -5 and 0.5. This will always work with polynomials with factors of ^2 in them.

If you have a TI-84 calculator or newer, there's a tool on it that will factor polynomials like this one for you just by giving it the numbers.

7 0

3 years ago

A norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. Find the dimensions of a norman

Yanka [14]

Answer:

$W\approx 8.72$ and $L\approx 15.57$ .

Step-by-step explanation:

Please find the attachment.

We have been given that a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. The total perimeter is 38 feet.

The perimeter of the window will be equal to three sides of rectangle plus half the perimeter of circle. We can represent our given information in an equation as:

$2L+W+\frac{1}{2}(2\pi r)=38$

We can see that diameter of semicircle is W. We know that diameter is twice the radius, so we will get:

$2L+W+\frac{1}{2}(2r\pi)=38$

$2L+W+\frac{\pi}{2}W=38$

Let us find area of window equation as:

$\text{Area}=W\cdot L+\frac{1}{2}(\pi r^2)$

$\text{Area}=W\cdot L+\frac{1}{2}(\pi (\frac{W}{2})^2)$

$\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W}{2})^2)$

$\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W^2}{4})$

$\text{Area}=W\cdot L+\frac{\pi}{8}W^2$

Now, we will solve for L is terms W from perimeter equation as:

$L=38-(W+\frac{\pi }{2}W)$

Substitute this value in area equation:

$A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2$

Since we need the area of window to maximize, so we need to optimize area equation.

$A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2$

$A=38W-W^2-\frac{\pi }{2}W^2+\frac{\pi}{8}W^2$

Let us find derivative of area equation as:

$A'=38-2W-\frac{2\pi }{2}W+\frac{2\pi}{8}W$

$A'=38-2W-\pi W+\frac{\pi}{4}W$

$A'=38-2W-\frac{4\pi W}{4}+\frac{\pi}{4}W$

$A'=38-2W-\frac{3\pi W}{4}$

To find maxima, we will equate first derivative equal to 0 as:

$38-2W-\frac{3\pi W}{4}=0$

$-2W-\frac{3\pi W}{4}=-38$

$\frac{-8W-3\pi W}{4}=-38$

$\frac{-8W-3\pi W}{4}*4=-38*4$

$-8W-3\pi W=-152$

$8W+3\pi W=152$

$W(8+3\pi)=152$

$W=\frac{152}{8+3\pi}$

$W=8.723210$

$W\approx 8.72$

Upon substituting $W=8.723210$ in equation $L=38-(W+\frac{\pi }{2}W)$ , we will get:

$L=38-(8.723210+\frac{\pi }{2}8.723210)$

$L=38-(8.723210+\frac{8.723210\pi }{2})$

$L=38-(8.723210+\frac{27.40477245}{2})$

$L=38-(8.723210+13.70238622)$

$L=38-(22.42559622)$

$L=15.57440378$

$L\approx 15.57$

Therefore, the dimensions of the window that will maximize the area would be $W\approx 8.72$ and $L\approx 15.57$ .

8 0

3 years ago