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2 years ago

11

Anna has 7.2 cups of cookie batter. If she uses all the batter to make 16 equal sized cookies, how much batter are in each cooki

e?

2 answers:

Lena [83]2 years ago

8 0

2.2 is the answer I got for that

lys-0071 [83]2 years ago

8 0

To get the answer, you must divide 7.2 by 16.
7.2/16=0.45
There are 0.45 cups of batter in each cookie.

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Which list shows these lengths in order from greatest to least?

finlep [7]

9514 1404 393

Answer:

√3/11, π/24, 3/25, 9/100

Step-by-step explanation:

The 2-decimal value of each of the numbers is ...

9/100 = 0.09

π/24 = 0.13

3/25 = 0.12

√3/11 = 0.16

Then the order from greatest to least is 0.16, 0.13, 0.12, 0.09. In the original form, that is ...

√3/11, π/24, 3/25, 9/100 . . . . . matches choice B

8 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

Evaluate the following expression will give branliest<br> 256 to the power of 5/8

Nana76 [90]

Answer:

32

Step-by-step explanation:

The formula needed to solve this question by hand is:

$x^{\frac{m}{n}} =\sqrt[n]{x^{m}}$

256^(5/8) = 8th root of 256^5

256^(5/8) = 8th root of 1280

<u>256^(5/8) = 32</u>

5 0

2 years ago

Read 2 more answers

What is the answer to the question above?

Bond [772]

Answer:

i beleive that B is the best option

4 0

2 years ago

Read 2 more answers

How would you do a,b,c,and d

I am Lyosha [343]

a) You are told the function is quadratic, so you can write cost (c) in terms of speed (s) as

... c = k·s² + m·s + n

Filling in the given values gives three equations in k, m, and n.

$28 = k\cdot 10^2+m\cdot 10+n\\21=k\cdot 20^2+m\cdot 20+n\\16=k\cdot 30^2+m\cdot 30+n$

Subtracting each equation from the one after gives

$-7=300k+10m\\-5=500k+10m$

Subtracting the first of these equations from the second gives

$2=200k\\\\k=\dfrac{2}{200}=0.01$

Using the next previous equation, we can find m.

$-5=500\cdot 0.01+10m\\\\m=\dfrac{-10}{10}=-1$

Then from the first equation

[tex]28=100\cdot 0.01+10\cdot (-1)+n\\\\n=37[tex]

There are a variety of other ways the equation can be found or the system of equations solved. Any way you do it, you should end with

... c = 0.01s² - s + 37

b) At 150 kph, the cost is predicted to be

... c = 0.01·150² -150 +37 = 112 . . . cents/km

c) The graph shows you need to maintain speed between 40 and 60 kph to keep cost at or below 13 cents/km.

d) The graph has a minimum at 12 cents per km. This model predicts it is not possible to spend only 10 cents per km.

4 0

3 years ago