<h2>
Therefore he took 40 gram of 
type solution and 10 gram of 
type solution.</h2>
Step-by-step explanation:
Given that , A pharmacist 13% alcohol solution another 18% alcohol solution .
Let he took x gram solution of type solution
and he took (50-x) gram of type solution.
Total amount of alcohol =![[x\times\frac{13}{100}] +[(50 -x) \times\frac{18}{100} ]](https://tex.z-dn.net/?f=%5Bx%5Ctimes%5Cfrac%7B13%7D%7B100%7D%5D%20%2B%5B%2850%20-x%29%20%5Ctimes%5Cfrac%7B18%7D%7B100%7D%20%5D)
gram
Total amount of solution = 50 gram
According to problem
⇔![\frac{ [x\times\frac{13}{100}] +[(50 -x) \times\frac{18}{100} ]}{50}= \frac{14}{100}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5Bx%5Ctimes%5Cfrac%7B13%7D%7B100%7D%5D%20%2B%5B%2850%20-x%29%20%5Ctimes%5Cfrac%7B18%7D%7B100%7D%20%5D%7D%7B50%7D%3D%20%5Cfrac%7B14%7D%7B100%7D)
⇔
⇔- 5x= 700 - 900
⇔5x = 200
⇔x = 40 gram
Therefore he took 40 gram of type solution and (50 -40)gram = 10 gram of type solution.
Answer:
$1440
Step-by-step explanation:
First, find the unit rate by dividing 300 by 5, which equals 60. So, Kendra earns $60 in one day. Then, multiply 60 by 24, which equals 1440. So, Kendra makes $1440 in 24 days. Hope this helped!
I believe it’s the second one because it’s the only one that has the same variable.
Answer:
The zeroes are -6, 1/2 and 2.
Step-by-step explanation:
f(x) = 2x3 + 7x2 - 28x + 12 = 0
From the first and last coefficient 2 and 12, one guess for a zero is x = 2.
So substituting x = 2:
f(2) = 16+ 28 - 56 + 12 = 0
So x = 2 is a zero and x - 2 is a factor of f(x)
Performing long division:
x - 2)2x3 + 7x2 - 28x + 12( 2x2 + 11x - 6 <------- Quotient
2x3 - 4x2
11x2 - 28x
11x2 - 22x
- 6x + 12
-6x + 12
.............
Now we solve
2x2 + 11x - 6 = 0
(2x - 1)(x + 6) = 0
2x - 1 = 0 or x + 6 = 0, so:
x = 1/2, x = -6.
Using synthetic division with a divisor zero value of 1/2, you get quotient coefficients of 4, 2, -4, -2, 2, 0. Dividing those by 2 (the coefficient of "a" in the given denominator), we get a quotient of
... 2x⁵ +x⁴ -2x³ -x² +x
You will get the same result using long division.
