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MArishka [77]
3 years ago
14

All the positive integers excluding O are known as​

Mathematics
2 answers:
TEA [102]3 years ago
4 0

Answer: Natural Numbers

Step-by-step explanation:

Ivahew [28]3 years ago
3 0
Natural numbers ..... since it’s not uncommon
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Sketch the image of quadrilateral PQRS under the following dilations:
kramer

Answer:

dam this been up for a long time

Step-by-step explanation:

its just too hard sorry man

7 0
3 years ago
He first five terms in the arithmetic sequence an = 8n, starting with n = 1.
Marat540 [252]
8,16,24,32,40
All you have to do is replace the n in 8n with the sequance number you're looking for
like for instance:
8(1)=8
8(2)=16
7 0
3 years ago
Read 2 more answers
Select the correct answer.
Elis [28]
C would be the right answer
8 0
2 years ago
Read 2 more answers
12х - 5 &gt;3х^2 +5<br> I am confused how to solve this.
Agata [3.3K]

Answer (3 different forms):

<u>1st Form:</u>

2+-\frac{1}{3} \sqrt{6}

<u>2nd Form:</u>

-\sqrt\frac{2}{3}+2

<u>3rd Form:</u>

\frac{6-\sqrt{6}}{3}

Or

(\frac{6-\sqrt{6}}{3}),(\frac{6+\sqrt{6}}{3})

Step-by-step explanation:

You can see that I put three different forms of answers, but they are all still equivalent answers. The reason why I got three different answers is because I put the inequality in three different calculators and they all got me different answers, but all three are still equivalent. The following three attachments are images of the three calculators that I used to solve.

Hope this helps and answers your question! :)

6 0
3 years ago
The total mass of the Sun is about 2×10^30 kg, of which about 76 % was hydrogen when the Sun formed. However, only about 12 % of
nignag [31]

Answer:

A. 1.8 ×10^{30} Kg

B i. 3.0 × 10^{17} seconds

  ii. 9.6 × 10^{9} years

C. After 9.2 × 10^{9} (9.2 billion) years

Step-by-step explanation:

Given that the mass of the Sun = 2× 10^{30} Kg.

Mass of hydrogen when Sun was formed = 76% of 2× 10^{30} Kg

                            = \frac{76}{100}  ×2× 10^{30} Kg

                           = 1.52 × 10^{30} Kg

Mass of hydrogen available for fusion = 12% of 1.52 × 10^{30} Kg

                           = \frac{12}{100} × 1.52 × 10^{30} Kg

                           = 1.824 ×10^{30} Kg

A. Total mass of hydrogen available for fusion over the lifetime of the sun is 1.8 ×10^{30} Kg.

B. Given that the Sun fuses 6 × 10^{11} Kg of hydrogen each second.

i. The Sun's initial hydrogen would last;

                                     \frac{1.8*10^{30} }{6*10^{11} }

                                 = 3.04 × 10^{17} seconds

The Sun's hydrogen would last 3.0 × 10^{17} seconds

ii. Since there are 31536000 seconds in a year, then;

The Sun's initial hydrogen would last;

                                     \frac{3.04*10^{17} }{31536000}

                                 = 9.640 × 10^{9} years

The Sun's hydrogen would last 9.6 × 10^{9} years.

C. Given that our solar system is now about 4.6 × 10^{9} years, then;

                               \frac{9.6*10^{9} }{4.6*10^{9} }

                             = 2.09

So that;   2 × 4.6 × 10^{9} = 9.2 × 10^{9} years

Therefore, we need to worry about the Sun running out of hydrogen for fusion after 9.2 × 10^{9} years.

6 0
3 years ago
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