These questions can be daunting at first, but they're pretty simple to solve.
First, we need to establish a common denominator. We have 2 / 3, 1 / 4, and
-4 / 3. The least common denominator we can get is by multiplying 4 and 3 together to get 12. So we will change the denominator as follows;
2 / 3, 1 / 4, -4 / 3 = 8 / 12, 3 / 12, -16 / 12
Now we can put these back into the equation.
8/12x + 3/12 = -16/12
8x + 3 = -16
It's simple math from here on out, but I'll show the process. What we can basically do now is take away the denominator because it doesn't matter now that it's common.
Subtract 3 from both sides. Now we have 8x = -19
Dividing by 8 on both sides of the equation will get you your answer.
x = -19/8
Hope this helps!
Answer:
Step-by-step explanation:
Let number of plastic containers collected by fourth grade= x
Then number of plastic containers collected by fifth grade students=x-216
OR
If number of plastic container collected by fifth grade is y
then number of plastic container collected by fourth grade=y+216
So, we can write it as follows
⇒ number of plastic containers collected by fourth grade= number of plastic containers collected by fifth grade +216
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