y + 3 = 
(x - 5)
the equation of a line in point- slope form is
y - b = m ( x - a )
where m is the slope and (a, b) a point on the line
here m = and (a,b) = (5 , - 3 )
y + 3 = (x - 5 ) ← in point-slope form
Answer: <u>6 is the slope and 4 is the y-intercept.</u>
Step-by-step explanation:
Based on the question, I feel this is the best way to answer your question. I'm assuming you are in a basic graphing class.
The best, most useful thing for you right now would be to learn slope-intercept form, y = mx + b, where m and b are constants.
Simply add 6x to both sides to get into this form
y = 6x + 4.
In slope-intercept form, m is the slope, and b is the y-intercept. Thus, 6 is the slope and 4 is the y-intercept.
Hope it helps and lmk if you need more <3 :)
Answer:
B={1,2,3,4,6and12}
n (B) =6
Step-by-step explanation:
<h3>Greetings !</h3>
Factor, in mathematics, a number or algebraic expression that divides another number or expression evenly—i.e., with no remainder. For example, 3 and 6 are factors of 12 because 12 ÷ 3 = 4 exactly and 12 ÷ 6 = 2 exactly. The other factors of 12 are 1, 2, 4, and 12.
Hope it helps!
Answer:
x = 6
Step-by-step explanation:
Since the triangle is equilateral then sides are congruent.
Equate any 2 sides and solve for x
13x + 5 = 8x + 35 ( subtract 8x from both sides )
5x + 5 = 35 ( subtract 5 from both sides )
5x = 30 ( divide both sides by 5 )
x = 6
JK = (13 × 6) + 5 = 78 + 5 = 83
KL = (17 × 6) - 19 = 102 - 19 = 83
JL = (8 × 6) + 35 = 48 + 35 = 83
ΔJKL is equilateral with side = 83
Answer:
x= -3 and y= 0
Step-by-step explanation:
5x+2y=-15
<u>2x-2y=-6 </u>
<u>7x =-21</u>
x= -3
Putting value of x in equation 1
5(-3) +2y=-15
-15+2y= -15
2y= 0
y= 0
This can be solved with the help of matrices
In matrix form the above equations can be written in the form
![\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%262%5C%5C2%26-2%5C%2F%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}x\\y\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-15%5C%5C-6%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Let
= A = X and = B
Then AX= B
or X= A⁻¹ B
where A⁻¹= adj A/ ║A║ where mod A≠ 0
adj A= ![\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26-2%5C%5C-2%265%5C%2F%5Cend%7Barray%7D%5Cright%5D)
║A║= ( 5*-2- 2*2)= -10-4= -14≠0
X= A⁻¹ B
=- 1/14
=- 1/14 ![\left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%2A-15%26%2B%20-2%2A-6%5C%5C-2%2A-15%26%2B%205%2A-6%5C%5C%5Cend%7Barray%7D%5Cright%5D)
=- 1/14 ![\left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%2030%26%2B12%5C%5C30%26%2B-30%5C%5C%5Cend%7Barray%7D%5Cright%5D)
=- 1/14 ![\left[\begin{array}{ccc}42\\0\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D42%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-42%2F14%5C%5C0%2F-14%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-3\\0\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
From here x= -3 and y= 0
Solution Set = [(-3,0)]
