Answer:
34 inches
Step-by-step explanation:
Area is base times height time half
Then plug in the values: 136= 0.5 x 8x
Then solve for x
Answer: width = 5 units
Step-by-step explanation:
Let L represent the length of the rectangle.
The width of a rectangle is the length minus 2 units. It means that the width of the rectangle is (L - 2) units.
The formula for determining the area of a rectangle is expressed as
Area = length × width
The area of the rectangle is 35 units. It means that
L(L - 2) = 35
L² - 2L = 35
L² - 2L - 35 = 0
L² + 5L - 7L - 35 = 0
L(L + 5) - 7(L + 5) = 0
L - 7 = 0 or L + 5 = 0
L = 7 or L = - 5
Since the length cannot be negative, then
L = 7 units
Width = 7 - 2 = 5 units
Answer:
![x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]](https://tex.z-dn.net/?f=x_3%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%263%261%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
According to the given situation, The computation of all x in a set of a real number is shown below:
First we have to determine the 
so that 
![\left[\begin{array}{cccc}1&-3&5&-5\\0&1&-3&5\\2&-4&4&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%5C0%261%26-3%265%5C%5C2%26-4%264%26-4%5Cend%7Barray%7D%5Cright%5D)
Now the augmented matrix is
![\left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\2&-4&4&-4\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%265%5C%20%7C%5C%200%5C%5C2%26-4%264%26-4%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
After this, we decrease this to reduce the formation of the row echelon
![R_3 = R_3 -2R_1 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&2&-6&6\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_3%20%3D%20R_3%20-2R_1%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%265%5C%20%7C%5C%200%5C%5C0%262%26-6%266%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_3 = R_3 -2R_2 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_3%20%3D%20R_3%20-2R_2%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%265%5C%20%7C%5C%200%5C%5C0%260%260%26-4%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_2 = 4R_2 +5R_3 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&4&-12&0\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_2%20%3D%204R_2%20%2B5R_3%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%264%26-12%260%5C%20%7C%5C%200%5C%5C0%260%260%26-4%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_2 = \frac{R_2}{4}, R_3 = \frac{R_3}{-4} \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&1\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_2%20%3D%20%5Cfrac%7BR_2%7D%7B4%7D%2C%20%20R_3%20%3D%20%5Cfrac%7BR_3%7D%7B-4%7D%20%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%260%5C%20%7C%5C%200%5C%5C0%260%260%261%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_1 = R_1 +3 R_2 \rightarrow \left[\begin{array}{cccc}1&0&-4&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_1%20%3D%20R_1%20%2B3%20R_2%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-4%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%260%5C%20%7C%5C%200%5C%5C0%260%260%26-1%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_1 = R_1 +5 R_3 \rightarrow \left[\begin{array}{cccc}1&0&-4&0\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_1%20%3D%20R_1%20%2B5%20R_3%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-4%260%5C%20%7C%5C%200%5C%5C0%261%26-3%260%5C%20%7C%5C%200%5C%5C0%260%260%26-1%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![x = \left[\begin{array}{c}4x_3&3x_3&x_3\\0\end{array}\right] \\\\ x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]](https://tex.z-dn.net/?f=x%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4x_3%263x_3%26x_3%5C%5C0%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%20x_3%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%263%261%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
By applying the above matrix, we can easily reach an answer
Answer:
28/55
Step-by-step explanation:
The probability of taking a red bead first and a white bead second is:
P(red, white) = (7/11) (4/10)
P(red, white) = 14/55
The probability of taking a white bead first and a red bead second is:
P(white, red) = (4/11) (7/10)
P(white, red) = 14/55
So the probability that one bead of each color is taken is:
P = P(red, white) or P(white, red)
P = 14/55 + 14/55
P = 28/55
Answer:
Latisha correctly verified that x² + x - 12 = (x - 3)(x + 4)
Step-by-step explanation:
In the problem, each student was to give their explanation of why the statement above is correct. In order to know if this statement is correct, you can also simply multiply the terms after the equal sign.
(x - 3)(x + 4)
Use foil method to multiply.
x² - 3x + 4x - 12
Combine like terms.
x² + x - 12
You can also find out if the statement is correct is multiplying and adding the numbers together.
-3 * 4 = -12
-3 + 4 = 1
So, Latisha is correct because her explanation verified the statement above.
