You're very close. However, the less steep line should go through (4,7) and not some point slightly above that location. Also, the shaded region should be above both dashed lines at the same time. So you won't include the portion that I've marked in blue (see attached). Other than that, it looks great.
Answer: $33.00
Step-by-step explanation:
Given
Mr. Claussen needs 15 Pounds of ground beef
It is stated that 2 pounds of beef cost $4.40
Using unitary method
1 Pound cost ![\frac{4.4}{2}=\$2.2](https://tex.z-dn.net/?f=%5Cfrac%7B4.4%7D%7B2%7D%3D%5C%242.2)
For 15 pounds it is
![\Rightarrow 15\times 2.2=\$33](https://tex.z-dn.net/?f=%5CRightarrow%2015%5Ctimes%202.2%3D%5C%2433)
Thus, he has to spend $33 for 15 pound meat
Answer:
- When we are having a rational expression i.e. a expression of the type:
![\dfrac{f(x)}{g(x)}](https://tex.z-dn.net/?f=%5Cdfrac%7Bf%28x%29%7D%7Bg%28x%29%7D)
Where f(x) and g(x) are polynomial functions.
Now the domain of this rational expression is whole of the real numbers except the points where the function g(x) will be zero.
Hence we have to exclude the points where the given denominator quantity is zero.
- Let us consider an example as:
Let R(x) denote the rational function as:
![R(x)=\dfrac{x^2}{(x-2)(x-3)}](https://tex.z-dn.net/?f=R%28x%29%3D%5Cdfrac%7Bx%5E2%7D%7B%28x-2%29%28x-3%29%7D)
Now the domain of this rational function will be whole of the real line minus the points where the denominator is zero.
We know that (x-2)(x-3) is zero when x=2 or x=3.
Hence, the domain of R(x) is: R- {2,3}.
Answer:
no solutions
Step-by-step explanation:
if the end result of solving an equation is untrue, it means there is no number that will satisfy the equation, therefore the equation will have no solutions
Answer:
-2/5
Step-by-step explanation:
y2-y1
-------- is -2/5 . after . u plug the 2 pairs
x2-x1