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Gelneren [198K]
4 years ago
11

What equation is ALWAYS TRUE? HELPPPPPP

Mathematics
1 answer:
nadezda [96]4 years ago
3 0

Answer:

2(x-1)=2x-2

Step-by-step explanation:

distribute the 2 into (x-1) so you get 2x-2 which is equal to 2x-2

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Shawn wants to model the number 13,450 using base-ten blocks. how many large cubes, flats, and longs does he need to model the n
enot [183]
1 rod or long is a column of 10 unit cubes.

10 longs placed side by side, form a flat (with dimensions 10×10)

10 flats, placed on top of each other form a large cube, with dimensions 10×10×10 = 1000 cubes.
-------------------------------------------------------------------------------------------------

each small cube (unit cube) represents the number 1.

thus each long represents 10
each flat represents 100

and each large cube represents 1,000


The number 13,450 can be written as

1*10,000+3*1000+4*100+5*10 using the expanded form.

This means that there are 13 large cubes, 4 flats and 5 longs needed to model the number.


Answer: 13 large cubes, 4 flats, 5 longs

5 0
4 years ago
What is the minimum product of two numbers whose difference is 36​? What are the​ numbers?
GalinKa [24]
Is suppose 36 and 0 would be the best.
36 - 0 = 36
36 x 0 = 0
Seems like that’s the answer
7 0
3 years ago
What is the LCM of 1/3 and 2/5
mariarad [96]
I hope this helps you

8 0
3 years ago
Solve for x. <br><br> 45(15x+20)−7x=56(12x−24)+6
Liula [17]
Greetings!

Solve for x.
45(15x+20)-7x=56(12x-24)+6
Distribute the Parenthesis.
675x+900-7x=56(12x-24)+6
675x+900-7x=672x-1344+6
Combine Like Terms.
668x+900=672x-1338
Add 1338 to both sides.
(668x+900)+1338=(672x-1338)+1338
Subtract 668x from both sides.
[(668x+900)+1338]-668x=[(672x-1338)+1338]-668x
Simplify.
900+1338=672x-668x
2238=4x
Divide both sides by 4.
(2238)/4=(4x)/4
Simplify.
559.5=x
x=559.5

Hope this helps.
-Benjamin

8 0
3 years ago
Read 2 more answers
A tank contains 100 kg of salt and 1000 L of water. A solution of a concentration 0.05 kg of salt per liter enters a tank at the
Umnica [9.8K]

Answer:

Step-by-step explanation:

Initial Quantity of salt Q(0) = 100 kg

Capacity of tank = 1000 L

Inflow of liquid = 10 l /minute and outflow = same rate thus leaving the content of volume 1000 L at any time

Assume after entering constantly mixes and drains out.

Concentration flow = inflow - outflow

= 0.05(10) - \frac{Q(t)}{1000} *10

i.e. Q'(t) = 0.5-0.01Q(t), Q(0) = 100

Q'(t) = -0.01(Q(t)-50), Q(0) = 100\\\frac{dQ}{Q-50} =-0.01 dt\\ln |Q-5| = -0.01t +C\\Q-50 = Ae^{-0.01t}

Use Q(0) = 5

-45 = A  

So the equation would be

Q-50 = -45 e^{-0.01t} \\Q(t) = 50-45 e^{-0.01t} \\

a) Initial concentratin = \frac{100}{1000} =0.10

b) Use the solution fo rDE substitute t =1

Q(1) = 50-45e^(-0.01) = 5.4477 kg

c) As t becomes large Q = 50

So concentration limit = 50/1000 = 0.05

8 0
3 years ago
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