1. (7 , 0) and (11 , 3) → B and D
substitute each coordinate point into the left side of the equation and if equal to 21 then it is a solution.
( 7 , 0) → (3 × 7) - ( 4× 0) = 21 - 0 = 21
(11 , 3) → (3 × 11 ) - ( 4 × 3) = 33 - 12 = 21
2. ( 2 , - 9 ) is a solution to the equation
substitute the coordinates into the equation and if equal to 13 the it is the solution.
(2 , - 9 ) → (5 × 2 ) - (-9)/3 = 10 + 9/3 = 10 + 3 = 13
(3 , - 6) → (5 × 3) - (- 6)/3 = 15 + 6/3 = 15 + 2 = 17
thus (2 , -9 ) is the solution
4. x- intercept = (- 45 , 0)
to find the x- intercept let y = 0
1/3 x + 0 = - 15
1/3x = - 15
multiply both sides by 3
x = 3 × - 15 = - 45
thus x -intercept = ( - 45 , 0)
Answer:
10
Step-by-step explanation:
Since m||n the sum of given angles is equal to 180 (they are supplementary)
6x + 10 + 10x + 10 = 180 add like terms
16x + 20 = 180 subtract 20 from from both sides
16x = 160 divide both sides by 16
x = 10
Answer:
The question needs more information but you can see that all the numbers in the 1st column are multiplied by 6 to equal the number in column 2
Step-by-step explanation:
3 x6 =18
6 x 6= 36
9 x6= 54
12 x6 =72
By definition of absolute value, you have

or more simply,

On their own, each piece is differentiable over their respective domains, except at the point where they split off.
For <em>x</em> > -1, we have
(<em>x</em> + 1)<em>'</em> = 1
while for <em>x</em> < -1,
(-<em>x</em> - 1)<em>'</em> = -1
More concisely,

Note the strict inequalities in the definition of <em>f '(x)</em>.
In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:


All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.
Answer:
T = 7M/N
Step-by-step explanation:
Cross multiply
NT = 7M
Divide by N
NT/N = 7M/N
T = 7M/N