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juin [17]
3 years ago
12

If 20+15=24 64+13=42 Then 11+28=?

Mathematics
1 answer:
Rus_ich [418]3 years ago
5 0
2+0+1+5x3= 24 and
6+4+1+3x3=42 than
<span>1+1+2+8x3=36

Final Answer 36 

:D Welcome</span>
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Bryce is making a model building. He raises the walls of the building by 2 centimeters, five times.
photoshop1234 [79]

Answer:

10 cm!

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Your middle school has 900 students one third of the students bring their lunch instead of buying your lunch at school what is t
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1/3 of all students bring lunch
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3 years ago
A rectangular parking area measuring 5000 ft squared is to be enclosed on three sides using​ chain-link fencing that costs ​$5.5
kaheart [24]

Answer:

Dimensions: 75.3778 ft and 66.3325 ft

Minimum price: $1658.31

Step-by-step explanation:

Let's call the length of the parking area 'x', and the width 'y'.

Then, we can write the following equations:

-> Area of the park:

x * y = 5000

-> Price of the fences:

P = 2*x*5.5 + y*5.5 + y*7

P = 11*x + 12.5*y

From the first equation, we have that y = 5000/x

Using this value in the equation for P, we have:

P = 11*x + 12.5*5000/x = 11*x + 62500/x

To find the minimum of this function, we need to take its derivative and then make it equal to zero:

dP/dx = 11 - 62500/x^2 = 0

x^2 = 65000/11

x = 250/sqrt(11) = 75.3778 ft

This is the x value that gives the minimum cost.

Now, finding y and P, we have:

x*y = 5000

y = 5000/75.3778 = 66.3325

P = 11*x + 62500/x = $1658.31

5 0
3 years ago
UREGNT -- BRAINLIEST!
Nady [450]

Answer:

12

Step-by-step explanation:

∛1728 = ∛2*2*2*2*2*2*3*3*3*3 = 2*2*3 = 12

3 0
3 years ago
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A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t)=(t)ln(5t). Find the acceleration of the par
lyudmila [28]

Answer:

a(\frac{1}{5e})=5e

Step-by-step explanation:

we are given equation for position function as

s(t)=tln(5t)

Since, we have to find acceleration

For finding acceleration , we will find second derivative

s'(t)=\frac{d}{dt}\left(t\ln \left(5t\right)\right)

=\frac{d}{dt}\left(t\right)\ln \left(5t\right)+\frac{d}{dt}\left(\ln \left(5t\right)\right)t

=1\cdot \ln \left(5t\right)+\frac{1}{t}t

s'(t)=\ln \left(5t\right)+1

now, we can find derivative again

s''(t)=\frac{d}{dt}\left(\ln \left(5t\right)+1\right)

=\frac{d}{dt}\left(\ln \left(5t\right)\right)+\frac{d}{dt}\left(1\right)

=\frac{1}{t}+0

a(t)=\frac{1}{t}

Firstly, we will set velocity =0

and then we can solve for t

v(t)=s'(t)=\ln \left(5t\right)+1=0

we get

t=\frac{1}{5e}

now, we can plug that into acceleration

and we get

a(\frac{1}{5e})=\frac{1}{\frac{1}{5e}}

a(\frac{1}{5e})=5e


5 0
3 years ago
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