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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: The answer would be D. 64

Step-by-step explanation: Understanding that the question noted that the second rectangle was dilated from PQRS, which has a perimeter of 16. With the coordinates of point P for the first rectangle being (2,0.5) whilst the second rectangle has point P at (8,2). To figure out the perimeter, divide point P from the second rectangle with point P from the first. The result would be 4. Thus, the scale factor is 4, which you then multiply the perimeter of PQRS by, which was 16. 16 times 4 equals 64.

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3 years ago
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The slope of KH is 1

The slope of GJ is -1

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Therefore, KH is perpendicular to GJ

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2 years ago
Determine a área total e perímetro da figura a seguir:<br><br> Apresente os cálculos.
ra1l [238]

Answer:

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2 years ago
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professor190 [17]

Answer:

P(x)=x^5+11x^3+28x

Step-by-step explanation:

<u>Roots of a polynomial</u>

If we know the roots of a polynomial, say x1,x2,x3,...,xn, we can construct the polynomial using the formula

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Where a is an arbitrary constant.

We know three of the roots of the degree-5 polynomial are:

x_1=0;\ x_2=\sqrt{7}\boldsymbol{i}:\ x_3=-2\boldsymbol{i}

We can complete the two remaining roots by knowing the complex roots in a polynomial with real coefficients, always come paired with their conjugates. This means that the fourth and fifth roots are:

x_4=-\sqrt{7}\boldsymbol{i}:\ x_3=+2\boldsymbol{i}

Let's build up the polynomial, assuming a=1:

P(x)=(x-0)(x-\sqrt{7}\boldsymbol{i})(x+\sqrt{7}\boldsymbol{i})(x-2\boldsymbol{i})(x+2\boldsymbol{i})

Since:

(a+b\boldsymbol{i})\cdot (a-b\boldsymbol{i})=a^2+b^2

P(x)=(x)(x^2+7)(x^2+4)

Operating the last two factors:

P(x)=(x)(x^4+11x^2+28)

Operating, we have the required polynomial:

\boxed{P(x)=x^5+11x^3+28x}

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Step-by-step explanation:

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