An actual Canadian quarter has a diameter of 23.88 mm. Calculate the scale factor used to create the
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The sum of angles in any quadrilateral, including trapezoid, is 360⁰.
Because we have <span>an isosceles trapezoid, we have 2 angles with measure 135⁰,
and we have 2 equal acute angles with measure x⁰.
So, we can find value of acute angle,
135*2 +2x =360⁰
270+2x=360
2x=360-270
2x=90
x=45⁰
So, acute angles in trapezoid = 45⁰.
From triangle ABC,
angle ACB =90⁰
angle A=45⁰,
so angle ABC= 180-(90-45)=45⁰
Triangle ABC is isosceles triangle,so |AC| = |CB|= 5 in.
So, longer base AA' = 5+4+5= 14 in
Now, we can find area of trapezoid.
shorter base = 4 in
longer base = 14 in
altitude =h = 5 in
Area of trapezoid =(1/2)(base1+base2)*h
Area of trapezoid = (1/2)(4+14)*5= 9*5=45 in²
Answer is 45 in².</span>
Because we have <span>an isosceles trapezoid, we have 2 angles with measure 135⁰,
and we have 2 equal acute angles with measure x⁰.
So, we can find value of acute angle,
135*2 +2x =360⁰
270+2x=360
2x=360-270
2x=90
x=45⁰
So, acute angles in trapezoid = 45⁰.
From triangle ABC,
angle ACB =90⁰
angle A=45⁰,
so angle ABC= 180-(90-45)=45⁰
Triangle ABC is isosceles triangle,so |AC| = |CB|= 5 in.
So, longer base AA' = 5+4+5= 14 in
Now, we can find area of trapezoid.
shorter base = 4 in
longer base = 14 in
altitude =h = 5 in
Area of trapezoid =(1/2)(base1+base2)*h
Area of trapezoid = (1/2)(4+14)*5= 9*5=45 in²
Answer is 45 in².</span>
Answer:
An equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be:
Step-by-step explanation:
We know that the slope-intercept form of the line equation is
y=mx+b
where m is the slope and b is the y-intercept.
Given the line
6x+y=2
Simplifying the equation to write into the slope-intercept form
y = -6x+2
So, the slope = -6
As we know that the slope of the perpendicular line is basically the negative reciprocal of the slope of the line.
Thus, the slope of the perpendicular line will be: -1/-6 = 1/6
Therefore, an equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be
substituting the values m = 1/6 and the point (6, -2)
subtract 2 from both sides
Therefore, an equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be: