Answer:
The equation ls: y = -4x - 16
Step-by-step explanation:
Given the point (-6, 8)
Given the line y = .25x - 7
Use a fraction:
y = 1/4x - 7
The slope is 1/4 (or .25)
The slope of a perpendicular line is the negative inverse or -4.
Use the point slope form and substitute:
y - y1 = m(x - x1)
y - 8 = -4(x - (-6))
y - 8 = -4x - 24
y = -4x - 16
Proof:
f(-6) = -4(-6) - 16
= 24 - 16
= 8, giving (-6, 8)
Obtuse, it does not fit Pythagoreans for a right triangle, two sides are not equal to be isosceles
Answer:
A=
Step-by-step explanation:
a−3=a1(5)
Step 1: Simplify both sides of the equation.
a−3=a1(5)
a+−3=5a
a−3=5a
Step 2: Subtract 5a from both sides.
a−3−5a=5a−5a
−4a−3=0
Step 3: Add 3 to both sides.
−4a−3+3=0+3
−4a=3
Step 4: Divide both sides by -4.
-4a/-4 = 3/-4
<h3>Your final answer is a=-3/4</h3>
Consider the provided phrase.
Four times the sum of five and a number
Let the number is n and the sum of five and a number can be written as:

Thus, four times the sum of five and a number can be written as:

Hence, the required expression is
.
What is phrase in math?
In mathematics, we may use a phrase to describe an expression. When we describe an expression in words that incorporates a variable, we are doing so by using an algebraic expression. An example of this may be the phrase "greater than" (an expression with a variable).
run the numbers. to conduct a rational analysis of an issue. You can do the math because I was the last person to depart yesterday night and the first person to arrive this morning, and since only the overweight security guy was present. on April 7, 2005, by Dave L. Flag.
Learn more about phrase brainly.com/question/7484425
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The curve

is parameterized by

so in the line integral, we have





You are mistaken in thinking that the gradient theorem applies here. Recall that for a scalar function

, we have gradient

. The theorem itself then says that the line integral of

along a curve

parameterized by

, where

, is given by

Specifically, in order for this theorem to even be considered in the first place, we would need to be integrating with respect to a vector field.
But this isn't the case: we're integrating

, a scalar function.