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iragen [17]
4 years ago
11

Yuri has a rope 12m long. he cuts it into pieces that are each 2m long. what fraction of the rope is one piece

Mathematics
2 answers:
Debora [2.8K]4 years ago
6 0
12 and divide bu 2 to get a answer of 6
TEA [102]4 years ago
5 0
Yuri will have 6 pieces because 12 divided by 2 is 6.


Hope that helped:)
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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
Y’all I need help I need a good grade on this
Andreyy89

Answer:

SLOPE IS -3

Step-by-step explanation:

4-7 over 0-(-1)

3 0
3 years ago
What is 11.45 as a fraction
Viefleur [7K]
11.45

the 11 is ur whole number...it stays ur whole number
the 0.45 is ur fraction...and since the last number (5) is in the hundredths place, put it over 100
so u now have : 11 45/100 which reduces to 11 9/20 and if u need it in improper fraction form, it is 229/20
6 0
3 years ago
Read 2 more answers
Make a function table to represent his total savings for 2,4,6,8 and 10 months.
OlgaM077 [116]
Well, let's say that bananas cost $1.60 per pound. That means that at step 0 the value is 0, at step 1 the value is$1.60, at step 2 the value is $$3.20 and so on. To make a table from this we'd need two columns and four rows. On the top row we right step followed by amount. In the step column, we list the numbers 0-2. In the amount column we list 0, $1.60, and $3.20. In the end, a function table looks like this:
5 0
4 years ago
(15 pts) 4. Find the solution of the following initial value problem: y"-10y'+25y = 0 with y(0) = 3 and y'(0) = 13
jolli1 [7]

Answer:

y(x)=3e^{5x}-2xe^{5x}

Step-by-step explanation:

The given differential equation is y''-10y'+25y=0

The characteristics equation is given by

r^2-10r+25=0

Finding the values of r

r^2-5r-5r+25=0\\\\r(r-5)-5(r-5)=0\\\\(r-5)(r-5)=0\\\\r_{1,2}=5

We got a repeated roots. Hence, the solution of the differential equation is given by

y(x)=c_1e^{5x}+c_2xe^{5x}...(i)

On differentiating, we get

y'(x)=5c_1e^{5x}+5c_2xe^{5x}+c_2e^{5x}...(ii)

Apply the initial condition y (0)= 3 in equation (i)

3=c_1e^{0}+0\\\\c_1=3

Now, apply the initial condition y' (0)= 13 in equation (ii)

13=5(3)e^{0}+0+c_2e^{0}\\\\13=15+c_2\\\\c_2=-2

Therefore, the solution of the differential equation is

y(x)=3e^{5x}-2xe^{5x}

5 0
3 years ago
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