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Shkiper50 [21]
3 years ago
10

The slope of the line that passes through the points (7,1) and (x,4). What is the value of w?

Mathematics
1 answer:
maks197457 [2]3 years ago
5 0

Answer:

  bbb

Step-by-step explanation:

bbb  idk sorry x

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(4-10)-(8 ÷(-2))

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If a triangle has sides of 8, 8, and 6, then it is classified as an _____ triangle.
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right

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I need help. Say two squirrels are mating in a tree and that tree is 56 feet tall. How many babies would they have if the sun wa
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4 0
3 years ago
Find a.<br> Round to the nearest tenth:<br> 2 cm<br> с<br> 150<br> 1050<br> a =<br> = [? ]cm
VLD [36.1K]

Answer:

a = 1.8 cm

Step-by-step explanation:

Here. we want to find the value of a

To do this, we are going to use the sine rule

it states that the ratio of a side and the sine of the angle that faces the side is a constant for all the sides of a triangle

From the diagram, we have that;

The angle facing a is;

180-15-105 = 60

So, we have that

a/sin 60 = 2/sin 105

a = 2sin 60/sin 105

a = 1.794 cm

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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