The equation is y = Three-halvesx + 1
Step-by-step explanation:
That line should be perpendicular to the given line : y-4 = -2/3 (x-6) and
It passes through (-2,-2)
Write the equation in slope-intercept form as;
y-4= -2/3 (x-6)
y-4=-2/3x + 4
y= -2/3x + 8
The slope m₁, of the equation is -2/3
For perpendicular lines, m₁*m₂= -1
So,finding m₂ where m₁ = -2/3
-2/3 * m₂ = -1
m₂ = -1 * - 3/2 = 3/2
The equation of the line passing through point (-2,-2) with m₂=3/2 will be;
y--2/x--2 = 3/2
y+2/x+2=3/2
2(y+2)=3(x+2)
2y+4=3x+6
2y=3x+6-4
2y=3x+2
y=3/2x +1 ⇒⇒Three-halves *x+1
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Equation of perpendicular lines : brainly.com/question/11482954
Keyword : equation, slope-intercept form, perpendicular
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Step-by-step explanation:
on line AB
X + 117 = 180 ( Angle sum property )
X = 63°
and
X+Y = 90° (given )
63° + Y = 90°
Y = 27°
When we approach limits, we are finding values that are infinitesimally approaching this x-value. Essentially, we consider the approximate location that this root or limit appears. This is essential when it comes to taking Calculus, and finding the limit or rate of change of a function.
When we are attempting limits questions, there are several tests we attempt first.
1. Evaluate the limit by substituting the value of the x-value as it approaches the value (direct evaluation of a limit)
2. Rearrangement of the function, such that we can evaluate the limit.
3. (TRIGONOMETRIC PROPERTIES)


4. Using L'Hopital's Rule for indeterminate limits, such as 0/0, -infinity/infinity, or infinity/infinity.
For example:
1)

We can do this using the first and second method.
<em>Method 1: Direct evaluation:</em>Substitute x = 0 to the function.


<em>Method 2: Rearranging the function
</em>We can see that x - 25 can be rewritten as: (√x - 5)(√x + 5)
By rewriting it in this form, the top will cancel with the bottom easily, and our limit comes out the same.



Every example works exactly the same way, and by remembering these criteria, every limit question should come out pretty naturally.
Step-by-step explanation:
f(x) = x² + x + 3/4
in general, such a quadratic function is defined as
f(x) = a×x² + b×x + c
the solution for finding the values of x where a quadratic function value is 0 (there are as many solutions as the highest exponent of x, so 2 here in our case)
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = 1
c = 3/4
x = (-1 ± sqrt(1² - 4×1×3/4))/(2×1) =
= (-1 ± sqrt(1 - 3))/2 = (-1 ± sqrt(-2))/2 =
= (-1 ± sqrt(2)i)/2
x1 = (-1 + sqrt(2)i) / 2
x2 = (-1 - sqrt(2)i) / 2
remember, i = sqrt(-1)
f(x) has no 0 results for x = real numbers.
for the solution we need to use imaginary numbers.
Answer:
y = 2/3x + 19/3
Step-by-step explanation:
First, this is your current equation: y = 2/3x + b. Plug in your point to the equation, it should look like this: 5 = 2/3(-2) + b. 2/3 times -2 equals -4/3. So, this is what your equation should look like: 5 = -4/3 + b. Add 4/3 to both sides of the equation to get 19/3 = b. Go back to your original equation and plug 19/3 to b. This is your final equation: y = 2/3x + 19/3. Hope this helped!