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Bumek [7]
1 year ago
5

How do I subtract 4m^2 + 2mn + 8n^2 from 2m^2 + 6mn + 2n^2 ?

Mathematics
1 answer:
SVETLANKA909090 [29]1 year ago
4 0
-2m^2+4mn-6n^2

1) Let's write out both expressions subtracting 4m²+2mn+8n² from 2m²+6mn+2n²

\begin{gathered} 2m^{2}+6mn+2n^{2}-(4m^{2}+2mn+8n^2)_{} \\ 2m^2+6mn+2n^2-4m^2-2mn-8n^2 \\ -2m^2+4mn-6n^2 \end{gathered}

2) Note that when we subtract 4m^2 + 2mn + 8n^2 from 2m^2 + 6mn + 2n^2 we need to swap the sign by placing -1 outside the parentheses and then combine like terms adding those terms algebraically.

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This means we increase the x value by 1.
velikii [3]

Answer:

The answer is option C, that is, (2,6)

6 0
3 years ago
I don’t get this can someone explain this
LekaFEV [45]

Answer:

15.7 units

Step-by-step explanation:

AB = 6, BC = 3

Therefore by Pythagoras theorem

AC = \sqrt{45}

perimeter \: of \:  \triangle \: ABC \\  = 6 + 3 +  \sqrt{45}  \\  = 9 + 6.70820393 \\  = 9 + 6.7 \\  = 15.7\: units

3 0
3 years ago
Class A has 20 pupils and class B has 19 pupils.
julsineya [31]

Answer:

47.5

Step-by-step explanation:

33+62=95

95/2=47.5

8 0
3 years ago
X+3/ x2-9x+20<br> A.4<br> B-3<br> C-4<br> D 5<br> E.3<br> F -5 <br> apex
serious [3.7K]

Answer:

  • local minimum at x = -3-2√14
  • local maximum at x = -3+2√14

Step-by-step explanation:

The first derivative simplifies to ...

  d/dx( ) = -(x^2 +6x -47)/(x^2 -9x +20)^2

This has zeros that can be found by the usual methods of solving quadratics:

  x = -3 ±2√14

The positive value of x corresponds to what I might call an "apex." (See the first attachment.) The negative value is where the function turns around an approaches the horizontal asymptote from below. (See the second attachment.)

3 0
3 years ago
A tank contains 200 liters of fluid in which 50 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu
mel-nik [20]

At the start, the tank contains A(0) = 50 g of salt.

Salt flows in at a rate of

(1 g/L) * (5 L/min) = 5 g/min

and flows out at a rate of

(A(t)/200 g/L) * (5 L/min) = A(t)/40 g/min

so that the amount of salt in the tank at time t changes according to

A'(t) = 5 - A(t)/40

Solve the ODE for A(t):

A'(t) + A(t)/40 = 5

e^(t/40) A'(t) + e^(t/40)/40 A(t) = 5e^(t/40)

(e^(t/40) A(t))' = 5e^(t/40)

e^(t/40) A(t) = 200e^(t/40) + C

A(t) = 200 + Ce^(-t/40)

Given that A(0) = 50, we find

50 = 200 + C  ==>  C = -150

so that the amount of salt in the tank at time t is

A(t) = 200 - 150 e^(-t/40)

7 0
4 years ago
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