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irakobra [83]
3 years ago
6

HELP ASAP!!!! 10 POINTS!!!

Mathematics
1 answer:
Kaylis [27]3 years ago
8 0

Answer:

sorry dont knowz

Step-by-step explanation:

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At a school there are five lessons in a day in total the five lessons last 4 hours assume that each session lasts the same amoun
ikadub [295]

Answer:

48 minutes long

Step-by-step explanation:

For a total amount of minutes the lessons are, take 4x60. 4 for the total time in the lessons and 60 for 60 minutes in an hour. 4x60= 240 minutes.

Take 240/5. 240 is the total number of minutes and with the five lessons being the same amount of time, you divide the total number of minutes by the number of lessons. 240/5= 48 minutes

7 0
2 years ago
I want solution...please help me
Art [367]

sin(<em>θ</em>) + cos(<em>θ</em>) = 1

Divide both sides by √2:

1/√2 sin(<em>θ</em>) + 1/√2 cos(<em>θ</em>) = 1/√2

We do this because sin(<em>x</em>) = cos(<em>x</em>) = 1/√2 for <em>x</em> = <em>π</em>/4, and this lets us condense the left side using either of the following angle sum identities:

sin(<em>x</em> + <em>y</em>) = sin(<em>x</em>) cos(<em>y</em>) + cos(<em>x</em>) sin(<em>y</em>)

cos(<em>x</em> - <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) - sin(<em>x</em>) sin(<em>y</em>)

Depending on which identity you choose, we get either

1/√2 sin(<em>θ</em>) + 1/√2 cos(<em>θ</em>) = sin(<em>θ</em> + <em>π</em>/4)

or

1/√2 sin(<em>θ</em>) + 1/√2 cos(<em>θ</em>) = cos(<em>θ</em> - <em>π</em>/4)

Let's stick with the first equation, so that

sin(<em>θ</em> + <em>π</em>/4) = 1/√2

<em>θ</em> + <em>π</em>/4 = <em>π</em>/4 + 2<em>nπ</em>   <u>or</u>   <em>θ</em> + <em>π</em>/4 = 3<em>π</em>/4 + 2<em>nπ</em>

(where <em>n</em> is any integer)

<em>θ</em> = 2<em>nπ</em>   <u>or</u>   <em>θ</em> = <em>π</em>/2 + 2<em>nπ</em>

<em />

We get only one solution from the second solution set in the interval 0 < <em>θ</em> < 2<em>π</em> when <em>n</em> = 0, which gives <em>θ</em> = <em>π</em>/2.

5 0
3 years ago
105.6 L/h = ___ L/min?
Drupady [299]
(\frac{105.6\ L}{hour})(\frac{1\ hour}{60\ minutes})

The hours cancel out and you divide 105.6 by 60 to get 

1.76 L/min
5 0
3 years ago
Use the method of lagrange multipliers to find
Yanka [14]

Answer:

a) The function is: f(x, y) = x + y.

The constraint is: x*y = 196.

Remember that we must write the constraint as:

g(x, y) = x*y - 196 = 0

Then we have:

L(x, y, λ) = f(x, y) +  λ*g(x, y)

L(x, y,  λ) = x + y +  λ*(x*y - 196)

Now, let's compute the partial derivations, those must be zero.

dL/dx =  λ*y + 1

dL/dy =  λ*x + 1

dL/dλ = (x*y - 196)

Those must be equal to zero, then we have a system of equations:

λ*y + 1 = 0

λ*x + 1 = 0

(x*y - 196) = 0

Let's solve this, in the first equation we can isolate  λ to get:

λ = -1/y

Now we can replace this in the second equation and get;

-x/y + 1 = 0

Now let's isolate x.

x = y

Now we can replace this in the last equation, and we will get:

(x*x - 196) = 0

x^2 = 196

x = √196 = 14

then the minimum will be:

x + y = x + x = 14 + 14 = 28.

b) Now we have:

f(x) = x*y

g(x) = x + y - 196

Let's do the same as before:

L(x, y, λ) = f(x, y) +  λ*g(x, y)

L(x, y, λ) = x*y +  λ*(x + y - 196)

Now let's do the derivations:

dL/dx = y + λ

dL/dy = x + λ

dL/dλ = x + y - 196

Now we have the system of equations:

y + λ = 0

x + λ = 0

x + y - 196 = 0

To solve it, we can isolate lambda in the first equation to get:

λ = -y

Now we can replace this in the second equation:

x - y = 0

Now we can isolate x:

x = y

now we can replace that in the last equation

y + y - 196 = 0

2*y - 196 = 0

2*y = 196

y = 196/2 = 98

The maximum will be:

x*y = y*y = 98*98 = 9,604

6 0
3 years ago
Please help !!
olga2289 [7]
If you have learned how to find the line of best fit manually, then you can do it that way. Perhaps you may want to just find a line that can connect at least two of the points and I believe that that line will be able to represent the other points because, in general, the points are pretty close to one another. 

If you don't want to do it manually and have a graphing calculator (which I recommend) then you can use that to find the line of best fit (and if you want then you can see how precise your points are with your r^2 value). Or there is a website (http://illuminations.nctm.org/Activity.aspx?id=4186), which you can use to help you to find the equation of that particular line. 

Once you have that done, then you can substitute 2009 for the x value in the equation and then see what y value the equation produces. That will then be your answer :) 
4 0
3 years ago
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