Dividing by a fraction is equivalent to multiply by its reciprocal, then:

Now, we need to express the quadratic polynomials using their roots, as follows:

where y1 and y2 are the roots.
Applying the quadratic formula to the first polynomial:
![\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B7%5Cpm%5Csqrt%5B%5D%7B%28-7%29%5E2-4%5Ccdot3%5Ccdot%28-6%29%7D%7D%7B2%5Ccdot3%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B7%5Cpm%5Csqrt%5B%5D%7B121%7D%7D%7B6%7D%20%5C%5C%20y_1%3D%5Cfrac%7B7%2B11%7D%7B6%7D%3D3%20%5C%5C%20y_2%3D%5Cfrac%7B7-11%7D%7B6%7D%3D-%5Cfrac%7B2%7D%7B3%7D%20%5Cend%7Bgathered%7D)
Applying the quadratic formula to the second polynomial:
![\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B-1%5Cpm%5Csqrt%5B%5D%7B1%5E2-4%5Ccdot2%5Ccdot%28-3%29%7D%7D%7B2%5Ccdot2%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B-1%5Cpm%5Csqrt%5B%5D%7B25%7D%7D%7B4%7D%20%5C%5C%20y_1%3D%5Cfrac%7B-1%2B5%7D%7B4%7D%3D1%20%5C%5C%20y_2%3D%5Cfrac%7B-1-5%7D%7B4%7D%3D-%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bgathered%7D)
Applying the quadratic formula to the third polynomial:
![\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B3%5Cpm%5Csqrt%5B%5D%7B%28-3%29%5E2-4%5Ccdot2%5Ccdot%28-9%29%7D%7D%7B2%5Ccdot2%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B3%5Cpm%5Csqrt%5B%5D%7B81%7D%7D%7B4%7D%20%5C%5C%20y_1%3D%5Cfrac%7B3%2B9%7D%7B4%7D%3D3%20%5C%5C%20y_2%3D%5Cfrac%7B3-9%7D%7B4%7D%3D-%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bgathered%7D)
Applying the quadratic formula to the fourth polynomial:
![\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B-1%5Cpm%5Csqrt%5B%5D%7B1%5E2-4%5Ccdot1%5Ccdot%28-2%29%7D%7D%7B2%5Ccdot1%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B-1%5Cpm%5Csqrt%5B%5D%7B9%7D%7D%7B2%7D%20%5C%5C%20y_1%3D%5Cfrac%7B-1%2B3%7D%7B2%7D%3D1%20%5C%5C%20y_2%3D%5Cfrac%7B-1-3%7D%7B2%7D%3D-2%20%5Cend%7Bgathered%7D)
Substituting into the rational expression and simplifying:
Good for Caleb, he better have bought cheese.
Triangle Area = .5 * base * height
48 = .5 * (x +4) * x
48 = (.5x + 2) * x
48 = .5x^2 + 2x
.5x^2 + 2x -48 = 0
x = 8
base = 12
Answer:
.
Step-by-step explanation:
Since repetition isn't allowed, there would be
choices for the first donut,
choices for the second donut, and
choices for the third donut. If the order in which donuts are placed in the bag matters, there would be
unique ways to choose a bag of these donuts.
In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type
times.
For example, if a bag includes donut of type
,
, and
, the count
would include the following
arrangements:
Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count
by
to find the actual number of donut combinations:
.
Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of
objects from a set of
distinct objects:
.
Given:
The expressions are:



To find:
The value of given expression by using integer tiles.
Solution:
We have,

Here, both number are positive. When we add 6 and 3 positive integer tiles, we get 9 positive integer tiles as shown in the below figure. So,

Similarly,

Here, 6 is positive and -4 is negative. It means we have 6 positive integer tiles and 4 negative integer tiles.
When we cancel the positive and negative integer tiles, we get 2 positive integer tiles as shown in the below figure. So,


Here, 6 is positive and -6 is negative. It means we have 6 positive integer tiles and 6 negative integer tiles.
When we cancel the positive and negative integer tiles, we get 0 integer tiles as shown in the below figure. So,

Therefore,
.