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3 years ago

Tessellations that use only one type of regular polygon are called semi-regular tessellations. A. True B. False

1 answer:

5 0

The answer is B... False. Tansellations that use only one type of regular polygon are not called semi-regular tessellations

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Let g(x) = |x + 1|, solve the equation g(x) = 2x + 4

hammer [34]

Answer:

Step-by-step explanation:

g(x)= -x-1 or g(x) = x+1

if g(x) = -x-1

-x-1=2x+4

-x=2x+5

-x/2=x+5

-x/2 -x=5

-x/2-2x/2=5

-x-2x=5

-3x=5

x=5/-3

if g(x) = x+1

x+1=2x+4

x=2x+3

x/2=x+3

x/2-x=3

x/2-2x/2=3

x-2x=3

-x=3

x=3

5 0

2 years ago

A company fills a warehouse will two types of goods A and B . they both come in tall boxes which cannot be stocked. one box of A

inessss [21]

Answer:

(1/2) * A + (1/2) * B <= 100; for A => 50; for B => 20

(5000) * A + (30000) * B <= 1500000; for A => 50; for B => 20

Step-by-step explanation:

There are two inequalities in mind, the first of the surface and the second of the price. Always bearing in mind that the minimum are 50 of A and 20 of B.

The first

A occupies 1/2 m and B occupies 1/2 m of surface, and the limit is 100 m of surface. Thus:

(1/2) * A + (1/2) * B <= 100; for A => 50; for B => 20

The second:

A costs 5,000 and B costs 30,000, and the limit is 1,500,000. Therefore:

(5000) * A + (30000) * B <= 1500000; for A => 50; for B => 20

5 0

3 years ago

A tank initially contains 60 gallons of brine, with 30 pounds of salt in solution. Pure water runs into the tank at 3 gallons pe

adoni [48]

Answer:

the amount of time until 23 pounds of salt remain in the tank is 0.088 minutes.

Step-by-step explanation:

The variation of the concentration of salt can be expressed as:

$\frac{dC}{dt}=Ci*Qi-Co*Qo$

being

C1: the concentration of salt in the inflow

Qi: the flow entering the tank

C2: the concentration leaving the tank (the same concentration that is in every part of the tank at that moment)

Qo: the flow going out of the tank.

With no salt in the inflow (C1=0), the equation can be reduced to

$\frac{dC}{dt}=-Co*Qo$

Rearranging the equation, it becomes

$\frac{dC}{C}=-Qo*dt$

Integrating both sides

$\int\frac{dC}{C}=\int-Qo*dt\\ln(\abs{C})+x1=-Qo*t+x2\\ln(\abs{C})=-Qo*t+x\\C=exp^{-Qo*t+x}$

It is known that the concentration at t=0 is 30 pounds in 60 gallons, so C(0) is 0.5 pounds/gallon.

$C(0)=exp^{-Qo*0+x}=0.5\\exp^{x} =0.5\\x=ln(0.5)=-0.693\\$

The final equation for the concentration of salt at any given time is

$C=exp^{-3*t-0.693}$

To answer how long it will be until there are 23 pounds of salt in the tank, we can use the last equation:

$C=exp^{-3*t-0.693}\\(23/60)=exp^{-3*t-0.693}\\ln(23/60)=-3*t-0.693\\t=-\frac{ln(23/60)+0.693}{3}=-\frac{-0.959+0.693}{3}= -\frac{-0.266}{3}=0.088$

5 0

2 years ago

1) Find 5 points on the line y = 3/5 x - 4 by picking 5 x values, plugging in, and getting 5 y values. You must pick x

Taya2010 [7]

It’s 5 for sure it is

4 0

2 years ago

800.5x(2x20to the 6 power

olasank [31]

Answer:

800.5×(2×20)=32020

Step-by-step explanation:

800.5×40=32020

3 0

3 years ago