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viktelen [127]
4 years ago
15

Draw a number line and show all numbers that are 2/3 of a unit to the left of 1/2

Mathematics
1 answer:
zubka84 [21]4 years ago
8 0

Answer:

See below.

Step-by-step explanation:

Draw a number line and draw points corresponding to -1, 0, and 1.

Mark 6 equal divisions between -1 and 0. Mark 6 equal divisions between 0 and 1.

Now you have the units between 1 and 0 and between 0 and -1 divided into sixths.

Now look at the point that corresponds to 1/2. It is the middle one-sixth mark between 0 and 1. Start there. That is 3/6 which is the same as 1/2.

You need to go 2/3 of a units to the left of 1/2.

2/3 = 4/6, so starting at 1/2, go 4 one-sixth marks to the left.

Now you are 2/3 unit to the left of 1/2.

See the number line below.

                -1                              0                              1

<----+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+------>

Start with a mark on 1/2, S in bold below.

                -1                              0             1/2             1

<----+---+---+---+---+---+---+---+---+---+---+---S---+---+---+---+---+------>

Now go 2/3 unit to left. 2/3 = 4/6, so go 4 one-sixth divisions to the left.

The point labeled with a bold E is the result. It is -1/6

                -1                      -1/6   0             1/2             1

<----+---+---+---+---+---+---+---E---+---+---+---S---+---+---+---+---+------>

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Answer:

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Hoped this helped! :)

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In the expansion of (1/ax +2ax^2)^5 the coefficient of x is five. Find the value of the constant a.
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Answer:

80x⁴

Step-by-step explanation:

(\frac{1}{ax} + 2ax^2)^5 = 5C_0(\frac{1}{ax})^5(2ax^2)^0 + 5C_1(\frac{1}{ax})^4(2ax^2)^1 + 5C_2(\frac{1}{ax})^3 (2ax^2)^2

                           + 5C_3 (\frac{1}{ax})^2(2ax^2)^3 + 5C_4(\frac{1}{ax})^1(2ax^2)^4 + 5C_5(\frac{1}{ax})^0(2ax^2)^5

5C_0(\frac{1}{ax})^5(2ax^2)^0  =1 \times (\frac{1}{ax})^5 \times 1 = \frac{1}{a^5x^5}\\\\5C_1(\frac{1}{ax})^4(2ax^2)^1  = 5 \times (\frac{1}{ax})^4 \times (2ax^2)^1 = 10 ax^2 \times \frac{1}{a^4x^4} = \frac{10}{a^3x^2}\\\\5C_2 (\frac{1}{ax})^3 (2ax^2)^2= 10 \times (\frac{1}{ax})^3 \times (2ax^2)^2 = 10 \times \frac{1}{a^3x^3} \times 4a^2x^4 = \frac{40x}{a}\\\\5C_3 (\frac{1}{ax})^2 (2ax^2)^3 = 10 \times (\frac{1}{ax})^2 \times (2ax^2)^3 = 10 \times \frac{1}{a^2x^2} \times 8a^3 x^6 = 80ax^4\\\\

5C_4(\frac{1}{ax})^1(2ax^2)^4 = 5 \times \frac{1}{ax} \times 16a^4x^8 = 80a^3x^7\\\\5C_5(\frac{1}{ax})^0(2ax^2)^5 = 1 \times 1 \times 32a^5x^{10}

The fourth term of the expansion has the constant a,

the coefficient of a is 80x⁴

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3 years ago
Jenny live 6 1/4 miles from school. Raymond lives 5 1/2 miles from school. How much farther from school does Jenny live then Ray
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6 1/4 - 5 1/2=
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