What is the value of |−44|?

2 answers:

8 0

Absolute value is simply the distance from zero on a number line. -44 is 44 away from zero, so the absolute value of -44 is

44

Natasha2012 [34]3 years ago

8 0

Answer:

Step-by-step explanation:

beacuse when on a number line l44l is 44 numbers away from 0 hope this helps :)

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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.

DedPeter [7]

Answer:

80x⁴

Step-by-step explanation:

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

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

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