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KiRa [710]
4 years ago
9

A cup got 1/4 of juice the 6 cups . how many days will it last

Mathematics
1 answer:
Elza [17]4 years ago
5 0
It would last 24 days.
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I Need Help Plz Need Answer Geometry Is Hard!!!
nekit [7.7K]

The line <em>AC</em> is a side that is common to both triangles.

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6 0
3 years ago
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Help me answer this question
Elden [556K]
Ok, so basically, the amount of orange pics is 3/5.
I think better w/ visuals, so i drew 5 circles, and colored 3 orange
Since the 3 circles represent 12 picks (because there are 12 picks total and 3/5 of the picks are orange) you nee to find out how many picks 1 circle represents.
In order to find this, do 12/4=3
One circle equals 3 picks
multiply 3 by 5, and you get 15
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7 0
3 years ago
Which graph represents this system?<br> first one is the system is it A.B.C or D
larisa86 [58]

Answer and Step-by-step explanation:

Well, Let's refer to the attachment shall we?

The first attachment is for y=2x+1

And the second attachment is for y=-4x+7

And the third attachment is for when you combine them together.

Now it's up to you which graph suites best? You will get your answer by referring it with my graph.

Hope this will help you!

3 0
4 years ago
The half-life of cesium-137 is 30 years. Suppose we have a
VMariaS [17]
(a) If y(t) is the mass (in mg) remaining after t years, then y(t) = y(0) e^{kt}=100e^{kt}.

y(30) = 100 e^{30k} = \frac{1}{2}(100) \implies e^{30k} = \frac{1}{2} \implies k = -(\ln 2) /30 \implies \\ \\&#10;y(t) = 100e^{-(\ln 2)t/30} = 100 \cdot 2^{-t/30}

(b) y(100) = 100 \cdot 2^{-100/30} \approx \text{9.92 mg}

(c)
100 e^{- (\ln 2)t/30} = 1\ \implies\ -(\ln 2) t / 30 = \ln \frac{1}{100}\ \implies\ \\ \\&#10;t = -30 \frac{\ln 0.01}{\ln 2} \approx \text{199.3 years}
 
3 0
3 years ago
g Farmer Violet wants to enclose one 990 square foot rectangular area of land using fencing. One side will use special fencing c
saw5 [17]

Answer:

Sides of the rectangular area:

x =  30  f

y =  33  f

Step-by-step explanation:

Let´s call   x  and y the sides of the rectangular area

A = x*y  = 990          y  = 990/x

The function cost is:

C = Costs ( sides x)  + costs ( sides y )

Cost of sides x    =  12*x +  10x

Cost of sides y    =  10*y  +  10 *y   =  2*10*y  = 20*y

C = 12*x + 10*x + 20*y =  22*x  + 20*y

The function cost as function of x is:

C(x)  = 22*x + 20*(990/y)  =  22*x  +  19800/x

Tacking derivative on both sides of the equation:

C´(x)  = 22  + [ - 19800/x²]

C´(x) = 0       22  - 19800/x²  =  0

Solving for x

22*x²  - 19800  =  0

22x²  =  19800

x² =  19800/22 =  900

x₁,₂ = ± 30        We dismiss negative root ( we never have negative lenghts)

Then     x =  30 f

And  y  =  990 / 30      y  = 33  

To see if x = 30 is a minimum for function C(x) we evaluate the second derivative.

C´´(x) = 2*x* 19800/x⁴      

C´´ (x) = 39600/x³    so C´´ is always greater than 0 then C (x) has a minimum at x = 30

Sides of the rectangular area are:

x  =  30

y  =  33

6 0
3 years ago
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