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GREYUIT [131]
3 years ago
11

The planet Mercury is approximately 6 × 10*7 miles from the sun . the distance between the sun and Mars is approximately 2 × 10*

8 .About how many times farther from the sun is Mars than Mercury ? A . 2 B . 3 C . 20 D . 30
​
Mathematics
2 answers:
vodomira [7]3 years ago
3 0
The answer is B

Because Mercury and the Sun are 60,000,00 miles apart and Mars is 200,000,000 miles away from the sun. Divide 200,000,000 and 60,000,000 and you will get 3.3 (Just round)
MAXImum [283]3 years ago
3 0

Answer:

B. 3

Step-by-step explanation:

Distance between Sun and Mercury = 6x10^7

Distance between Sun and Mars = 2x10^8

Assuming Mars is x times farther from the Sun, implies

(6x10^7)x = 2x10^8

therefore x = (2x10^8)/(6x10^7)

                   Eliminating common factors( 2x10^7) is common

therefore x = (1x10)/3

                x= 3.33

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Veseljchak [2.6K]

Dividing by a fraction is equivalent to multiply by its reciprocal, then:

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Now, we need to express the quadratic polynomials using their roots, as follows:

ay^2+by+c=a(y-y_1)(y-y_2)

where y1 and y2 are the roots.

Applying the quadratic formula to the first polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}

Applying the quadratic formula to the second polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}

Applying the quadratic formula to the third polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}

Applying the quadratic formula to the fourth polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}

Substituting into the rational expression and simplifying:

\begin{gathered} \frac{3(y-3)(y+\frac{2}{3})2(y-1)(y+\frac{3}{2})}{2(y-3)(y+\frac{3}{2})(y-1)(y+2)}= \\ =\frac{3(y+\frac{2}{3})}{2(y+2)}= \\ =\frac{3y+2}{2y+4} \end{gathered}

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1 year ago
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aleksandrvk [35]

Answer:

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Notice that the segment that joins the office store with the art gallery, has a length that equal the distance between the art gallery and the bank, plus the distance between the bank and the office supply store. That is;

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7 0
3 years ago
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natita [175]

Answer:

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Step-by-step explanation:

Note that the circumcentre is equally distant from the triangle's 3 vertices.

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2x - 15 = 7 ( add 15 to both sides )

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Answer:

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<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
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<u>Step 3: Redefine Systems</u>

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<u>Step 4: Solve for </u><em><u>y</u></em>

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  1. Combine 2 equations:                    11y = -44
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5 0
3 years ago
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Alja [10]

Answer:

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