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mr Goodwill [35]
3 years ago
10

The mass of the sun is 2.13525×1030 kilograms. The mass of Mercury is 3.285×1023 kilograms. How many times greater is the mass o

f the sun than the mass of Mercury? Drag and drop the values into the boxes to represent the answer in scientific notation.
Mathematics
2 answers:
Helga [31]3 years ago
8 0
<span>6.5 x 10^6 To answer this question, you need to divide the mass of the sun by the mass of mercury. So 2.13525 x 10^30 / 3.285 x 10^23 = ? To do the division, divide the mantissas in the normal fashion 2.13525 / 3.285 = 0.65 And subtract the exponents. 30 - 23 = 7 So you get 0.65 x 10^7 Unless the mantissa is zero, the mantissa must be greater than or equal to 0 and less than 10. So multiply the mantissa by 10 and then subtract 1 from the exponent, giving 6.5 x 10^6 So the sun is 6.5 x 10^6 times as massive as mercury.</span>
Rzqust [24]3 years ago
7 0

Answer:

6.5 x 10∧6 is the answer.

Step-by-step explanation:

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answer

Step-by-step explanation:

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Kryger [21]

Check the picture below.

\bf \stackrel{\measuredangle DAC}{4y+2x}~~=~~\stackrel{\measuredangle BCA}{9y-x}\implies 2x=5y-x\implies 3x=5y\implies \boxed{x=\cfrac{5y}{3}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\measuredangle DAB}{[(4y+2x)+35]}~~+~~\stackrel{\measuredangle ADC}{5x+4}~~=~~180\implies 4y+2x+5x+39 = 180 \\\\\\ 4y+7x+39=180\implies 4y+7x=141\implies \stackrel{\textit{substituting "x"}}{4y+7\left( \boxed{\cfrac{5y}{3}} \right)} = 141

\bf 4y+\cfrac{35y}{3}=141\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3\left( 4y+\cfrac{35y}{3} \right)=3(141)}\implies 12y+35y=423 \\\\\\ 47y=423\implies y=\cfrac{423}{47}\implies \blacktriangleright y = 9 \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{since we know that}}{x=\cfrac{5y}{3}}\implies x = \cfrac{5(9)}{3}\implies \blacktriangleright x = 15 \blacktriangleleft

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4 years ago
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