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raketka [301]
3 years ago
9

2x^2+15x+2=0 Need the answers for the two x's that are equal to 0

Mathematics
1 answer:
xxTIMURxx [149]3 years ago
6 0

Answer:

x = -2

Step-by-step explanation:

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Identify the coefficient(s) of the variable in the expression below.
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D

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25 and 6

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expeople1 [14]

Answer:

See below

Step-by-step explanation:

1) \:  \frac{1}{2}  \times 5 \times 6 = 15 \:  {cm}^{2}  \\  \\ 2) \:  \frac{1}{2}  \times 10 \times 6 = 30 \:  {cm}^{2}   \\  \\ 3) \: 5 \times 3 = 15 \:  {cm}^{2}  \\  \\ 4) \: 10 \times 6= 60\:  {cm}^{2}

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3 years ago
The constraints of a problem are listed below. What are the vertices of the feasible region?<br>​
Mamont248 [21]

Answer:

Option 4 : (0.\frac{3}{2} ) \ , \ (0,2) \ , \ (6,0) \ , \ (\frac{9}{4} ,0)

Step-by-step explanation:

<u>See the attached figure:</u>

To find the vertices of the feasible region, we need to graph the constraints, then find the area included by them, then calculate the vertices which is the intersection between each two of them.

As shown, the shaded area represents the solution of the constraints

So, the vertices of the feasible region are:

(0.\frac{3}{2} ) \ , \ (0,2) \ , \ (6,0) \ , \ (\frac{9}{4} ,0)

8 0
3 years ago
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Lubov Fominskaja [6]
ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta \ \textless \  0\ then\ no\ solution\\\\if\ \Delta =0\ then\ one\ solution\ x_0=\dfrac{-b}{2a}\\\\if\ \Delta \ \textgreater \  0\ then\ two\ solutions\ x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\-----------------------------

x^2+20x+98=0\\a=1;\ b=20;\ c=98\\\\\Delta=20^2-4\cdot1\cdot98=400-392=8 \ \textgreater \  0\\\sqrt\Delta=\sqrt8=\sqrt{4\cdot2}=\sqrt4\cdot\sqrt2=2\sqrt2\\\\x_1=\dfrac{-20-2\sqrt2}{2\cdot1}=\dfrac{-20-2\sqrt2}{2}=-10-\sqrt2\\\\x_2=\dfrac{-20+2\sqrt2}{2\cdot1}=\dfrac{-20+2\sqrt2}{2}=-10+\sqrt2
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3 years ago
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