30043——————3.0043 x 10^4
30.043—————— 3.0043 x10^1
0.0030043—————-3.0043 x 10^-3
0.30043—————-3.0043 x 10^-1
3.0043————-3.0043 x 10^0
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Answer:
Step-by-step explanation:
195xy is ur lowest common denominator
The solution of the equation is x = 71 / 36.
<h3>How to solve an equation with one variable</h3>
Herein we have an equation with several rational constants and only one variable, x, which has to be cleared by using algebra procedures. The procedure is shown below:
2 · x + 1 / 3 + x - 1 / 4 = 13 / 2 Given
2 · x + x = 13 / 2 - 1 / 3 - 1 / 4 Compatibility with addition / Existence of additive inverse / Modulative property
3 · x = 71 / 12 Definitions of addition and subtraction / Distributive property
x = 71 / 36 Compatibility with multiplication / Existence of multiplicative inverse / Modulative property / Result
The solution of the equation is x = 71 / 36.
To learn more on equations: brainly.com/question/10413253
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THIS IS THE COMPLETE QUESTION BELOW
The demand equation for a product is p=90000/400+3x where p is the price (in dollars) and x is the number of units (in thousands). Find the average price p on the interval 40 ≤ x ≤ 50.
Answer
$168.27
Step by step Explanation
Given p=90000/400+3x
With the limits of 40 to 50
Then we need the integral in the form below to find the average price
1/(g-d)∫ⁿₐf(x)dx
Where n= 40 and a= 50, then if we substitute p and the limits then we integrate
1/(50-40)∫⁵⁰₄₀(90000/400+3x)
1/10∫⁵⁰₄₀(90000/400+3x)
If we perform some factorization we have
90000/(10)(3)∫3dx/(400+3x)
3000[ln400+3x]₄₀⁵⁰
Then let substitute the upper and lower limits we have
3000[ln400+3(50)]-ln[400+3(40]
30000[ln550-ln520]
3000[6.3099×6.254]
3000[0.056]
=168.27
the average price p on the interval 40 ≤ x ≤ 50 is
=$168.27