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Hatshy [7]
2 years ago
10

The body of a 154-pound person contains approximately 2×10^2 milligrams of gold and 6×10^3 milligrams of aluminum. Based on this

information, the number of milligrams of aluminum in the body is how many times greater than the number of milligrams of gold in the body?
can you show your work too?
re asked it cs some bozo stole my points but pls hellp asap
Mathematics
1 answer:
hram777 [196]2 years ago
3 0

Answer:

2x10^2 and 6x 10^

Step-by-step explanation:

Hope I helped UwU

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What form is this equation in y= -3(x-5)^2+7 in?
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vertex form

Step-by-step explanation:

y= -3(x-5)^2+7

This equation is written in vertex form

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8 0
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The sum of a number and 4.7 is 8.1.
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<em>Greetings from Brasil...</em>

Be that unknown number X

Then we can write the following expression:

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3 years ago
The amounts of food waste generated in a region during two years were 37,500,000 tons and 30,400,000 tons. What was the total fo
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3 0
2 years ago
Solve. Good luck! Please do not try to google this.
Elena L [17]
\frac{x^{2} + x - 2}{6x^{2} - 3x} = \sqrt{2x} + \frac{3x^{2}}{2}
\frac{x^{2} + 2x - x - 2}{3x(x) - 3x(1)} = \frac{2\sqrt{2x}}{2} + \frac{3x^{2}}{2}
\frac{x(x) + x(2) - 1(x) - 1(2)}{3x(x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}
\frac{x(x + 2) - 1(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}
\frac{(x - 1)(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}
\frac{(x - 1)(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}
3x(2x - 1)(2\sqrt{2x} + 3x^{2}) = 2(x + 2)(x - 1)
3x(2x(2\sqrt{2x} + 3x^{2}) - 1(2\sqrt{2x} + 3x^{2}) = 2(x(x - 1) + 2(x - 1))
3x(2x(2\sqrt{2x}) + 2x(3x^{2}) - 1(2\sqrt{2x}) - 1(3x^{2})) = 2(x(x) - x(1) + 2(x) - 2(1)
3x(4x\sqrt{2x} + 6x^{3} - 2\sqrt{2x} - 3x^{2}) = 2(x^{2} - x + 2x - 2)
3x(4x\sqrt{2x} - 2\sqrt{2x} + 6x^{3} - 3x^{2}) = 2(x^{2} + x - 2)
3x(4x\sqrt{2x}) - 3x(2\sqrt{2x}) + 3x(6x^{3}) - 3x(3x^{2}) = 2(x^{2}) + 2(x) - 2(2)
12x^{2}\sqrt{2x} - 6x\sqrt{2x} + 18x^{4} - 9x^{3} = 2x^{2} + 2x - 4
12x^{2}\sqrt{2x} - 6x\sqrt{2x} = -18x^{4} + 9x^{3} + 2x^{2} + 2x - 4
6x\sqrt{2x}(2x) - 6x\sqrt{2x}(1) = -9x^{3}(2x) - 9x^{3}(-1) + 2(x^{2}) + 2(x) - 2(2)
6x\sqrt{2x}(2x - 1) = -9x^{3}(2x - 1) + 2(x^{2} + x - 2)
5 0
3 years ago
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