Answer with explanation:
1. The given equations are
3x -5 y=2
-x+2 y= 0
⇒The matrix in the form of , AX=B, is
![A=\left[\begin{array}{cc}3&-5\\-1&2\end{array}\right] ,\\\\ X=\left[\begin{array}{c}x&y\end{array}\right],\\\\B=\left[\begin{array}{c}2&0\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%26-5%5C%5C-1%262%5Cend%7Barray%7D%5Cright%5D%20%2C%5C%5C%5C%5C%20X%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%2C%5C%5C%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%260%5Cend%7Barray%7D%5Cright%5D)
Adj.A=Transpose of cofactor of Matrix A
![Adj.A=\left[\begin{array}{cc}2&1\\5&3\end{array}\right] ,\\\\ |A|=6-5\\\\|A|=1\\\\\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{cc}2&5\\1&3\end{array}\right] \times \left[\begin{array}{c}2&0\end{array}\right]\\\\x=4, y=2](https://tex.z-dn.net/?f=Adj.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%261%5C%5C5%263%5Cend%7Barray%7D%5Cright%5D%20%2C%5C%5C%5C%5C%20%7CA%7C%3D6-5%5C%5C%5C%5C%7CA%7C%3D1%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%265%5C%5C1%263%5Cend%7Barray%7D%5Cright%5D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%260%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5Cx%3D4%2C%20y%3D2)
2.
The given equations are
x+y-z=2
x+z=7
2 x +y+z=13
⇒The matrix in the form of , AX=B, is
![A=\left[\begin{array}{ccc}1&1&-1\\1&0&1\\2&1&1\end{array}\right]\\\\ X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\\\\B= \left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B\\\\a_{11}=-1,a_{12}=1,a_{13}=1,a_{21}=-2,a_{22}=3,a_{23}=1,a_{31}=1,a_{32}=-2,a_{33}=-1\\\\|A|=1\times(0-1)-1\times(1-2)-1\times(1-0)\\\\=-1+1-1\\\\|A|=-1\\\\Adj.A=\left[\begin{array}{ccc}-1&-2&1\\1&3&-2\\1&1&-1\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%26-1%5C%5C1%260%261%5C%5C2%261%261%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%20X%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5CB%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5C%5C7%5C%5C13%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5Crightarrow%20X%3DA%5E%7B-1%7DB%5C%5C%5C%5C%5Crightarrow%20X%3D%5Cfrac%7BAdj.A%7D%7B%7CA%7C%7D%5Ctimes%20B%5C%5C%5C%5Ca_%7B11%7D%3D-1%2Ca_%7B12%7D%3D1%2Ca_%7B13%7D%3D1%2Ca_%7B21%7D%3D-2%2Ca_%7B22%7D%3D3%2Ca_%7B23%7D%3D1%2Ca_%7B31%7D%3D1%2Ca_%7B32%7D%3D-2%2Ca_%7B33%7D%3D-1%5C%5C%5C%5C%7CA%7C%3D1%5Ctimes%280-1%29-1%5Ctimes%281-2%29-1%5Ctimes%281-0%29%5C%5C%5C%5C%3D-1%2B1-1%5C%5C%5C%5C%7CA%7C%3D-1%5C%5C%5C%5CAdj.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26-2%261%5C%5C1%263%26-2%5C%5C1%261%26-1%5Cend%7Barray%7D%5Cright%5D)
![\frac{Adj.A}{|A|}=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\\\\X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\times\left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\x=3,y=3,z=4](https://tex.z-dn.net/?f=%5Cfrac%7BAdj.A%7D%7B%7CA%7C%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%26-1%5C%5C-1%26-3%262%5C%5C-1%26-1%261%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5CX%3DA%5E%7B-1%7DB%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%26-1%5C%5C-1%26-3%262%5C%5C-1%26-1%261%5Cend%7Barray%7D%5Cright%5D%5Ctimes%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5C%5C7%5C%5C13%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5Cx%3D3%2Cy%3D3%2Cz%3D4)
The answers to each of the given problems are;
1) Sum of two smallest integers = 23
2) 3x + 6
3) 7.5 m/s²
<h3>How to find the sum of consecutive Integers?</h3>
1) We are told that sum of 4 consecutive integers is increased by 20 and equals 70. Thus, we have;
x + (x + 1) + (x + 2) + (x + 3) + 20 = 70
4x + 26 = 70
4x = 70 - 26
4x = 44
x = 44/4
x = 11
Thus, sum of two smallest integers = 11 + 11 + 1 = 23
2) Let the consecutive odd numbers be;
x, (x + 2) and (x + 4)
Sum of these consecutive odd numbers is;
x + x + 2 + x + 4 = 3x + 6
3) We are given the equation to find the acceleration as;
(v_final)² - (v_initial)² = 2ad
We are given;
v_final = 40 m/s
v_initial = 10 m/s
d = 100 m
Thus;
40² - 10² = 2a(100)
1500 = 200a
a = 1500/200
a = 7.5 m/s²
Read more about sum of integers at; brainly.com/question/17695139
#SPJ1
y -3 = 2(x-4)
Use distributive property on the right side:
y - 3 = 2x -8
Add 3 to both sides:
y = 2x -5
The answer is A.
Answer:
Step-by-step explanation:
Method 1: Using a calculator <em>instead</em> of the unit circle
The unit circle gives coordinates pairs for the <em>cos</em> and <em>sin</em> values at a certain angle. Therefore, if an angle is given, use a calculator to evaluate the functions at cos(angle) and sin(angle).
Method 2: Using the unit circle
Use the unit circle to locate the angle measure of 150° (or 5π/6 radians) and use the coordinate pair listed by the value (see attachment).
This coordinate pair is (-√3/2, 1/2).
Answer:
6+12b
Step-by-step explanation:
i took a test with this question and got it right
