D
Sq rt of 8 is sq rt of 4 times sq rt of 2
✿————✦————✿————✦————✿
The answer is: <u>2(k2−4k)(2c+5)</u>
✿————✦————✿————✦————✿
Step:
* Consider 2ck2+5k2−8ck−20k. Do the grouping 2ck2+5k2−8ck−20k=(2ck2+5k2) +(−8ck−20k), and factor out k2 in the first and −4k in the second group.
* Factor out the common term 2c+5 by using the distributive property.
* Rewrite the complete factored expression.
✿————✦————✿————✦————✿
Answer:
$47.50
Step-by-step explanation:
You just take $380 and divide it by 8 hours to get the amount of money per hour, which is $47.50 .
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
Using the sum (addition) and two 2-digit addends, 12 + 34 can be answered without regrouping.
