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photoshop1234 [79]
3 years ago
12

5y - 31 Р 6x - 2y 2y + 5 4x + 4 T R S 7x - 17

Mathematics
1 answer:
tino4ka555 [31]3 years ago
5 0

you take each other you do multiplication and you have answer

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A mole of a chemical element contains approximately 6.02 × 1023 atoms. Match the following moles of elements with the approximat
makvit [3.9K]

Answer:

9.03x10^16---------1.5x10^-7 helium

2.709x10^24-------4.5 nitrogen

1.9264x10^28-------3.2x10^4 hydrogen

4.8782x10^14--------8.1x10^-10 xenon

Step-by-step explanation: i got it right PERIOD PURRRRR

3 0
2 years ago
Read 2 more answers
Identify the radius and center.<br><br> x^2 + y^2 - 2x + 4y - 11 = 0
miss Akunina [59]
<h2>Hello!</h2>

The answer is:

Center: (1,-2)

Radius: 4 units.

<h2>Why?</h2>

To solve the problem, using the given formula of a circle, we need to find its standard equation form which is equal to:

(x-h)^{2}+(y-k)^{2}=r^{2}

Where,

"h" and "k"are the coordinates of the center of the circle and "r" is its radius.

So, we need to complete the square for both variable "x" and "y".

The given equation is:

x^2+y^2-2x+4y-11=0

So, solving we have:

x^2+y^2-2x+4y=11

(x^2-2x+(\frac{2}{2})^{2} )+(y^2+4y+(\frac{4}{2})^{2})=11+(\frac{2}{2})^{2} +(\frac{4}{2})^{2}\\\\(x^2-2x+1)+(y^2+4y+4)=11+1+4\\\\(x^2-1)+(y^2+2)=16

(x^2-1)+(y^2-(-2))=16

Now, we have that:

h=1\\k=-2\\r=\sqrt{16}=4

So,

Center: (1,-2)

Radius: 4 units.

Have a nice day!

Note: I have attached a picture for better understanding.

5 0
3 years ago
Simplify:
OverLord2011 [107]

Step-by-step explanation:

Simplify :

x4−6x3+2x−7+7x3−x+5x2+2−x4

x4−6x3+2x−7+7x3−x+5x2+2−x4

Rearranging and collecting the like terms, we get:

=(x4−x4)+(−6x3+7x3)+5x2+(2x−x)+(−7+2)

=(1−1)x4+(−6+7)x3+5x2+(2−1)x+(−7+2)

=(1)x3+5x2+(1)x−7+2

=x3+5x2+x−5

8 0
3 years ago
Use the graph to answer problem
Alina [70]

Answer:

The option you chose is the right answer which is (0,-9).

4 0
3 years ago
78 divided by 3 * 8 + 9 - 7 =
tester [92]

Answer:

210

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
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