Answer:
27000
Step-by-step explanation:
Answer:
the LCM is 6x^2(x-1)^2(x+2)
Answer:
(d) m∠AEB = m∠ADB
Step-by-step explanation:
The question is asking you to compare the measures of two inscribed angles. Each of the inscribed angles intercepts the circle at points A and B, which are the endpoints of a diameter.
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<h3>applicable relations</h3>
Several relations are involved here.
- The measures of the arcs of a circle total 360°
- A diameter cuts a circle into two congruent semicircles
- The measure of an inscribed angle is half the measure of the arc it intercepts
<h3>application</h3>
In the attached diagram, we have shown inscribed angle ADB in blue. The semicircular arc it intercepts is also shown in blue. A semicircle is half a circle, so its arc measure is half of 360°. Arc AEB is 180°. That means inscribed angle ADB measures half of 180°, or 90°. (It is shown as a right angle on the diagram.)
If Brenda draws angle AEB, it would look like the angle shown in red on the diagram. It intercepts semicircular arc ADB, which has a measure of 180°. So, angle AEB will be half that, or 180°/2 = 90°.
The question is asking you to recognize that ∠ADB = 90° and ∠AEB = 90° have the same measure.
m∠AEB = m∠ADB
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<em>Additional comment</em>
Every angle inscribed in a semicircle is a right angle. The center of the semicircle is the midpoint of the hypotenuse of the right triangle. This fact turns out to be useful in many ways.
The equation would be something like this:
y=72.39+ .18x+.06x
then simplify
y=72.39+.24x
Answer:
See explanation
Step-by-step explanation:
The place to begin in attempting to solve a quadratic equation is to try to factorize it. If it can not be factorized, then look at the side of equation that contains the variable (x,b,m,n and so on). If that side is a perfect square, try to take square root of both sides.
Also, if the coefficient of the squared term in the equation is 1 and the coefficient of the middle term bearing the variable is an even number, then you can attempt to use completing the square method. Formula method can be adopted as a last resort.
x^2 -4x + 4 =0 can be solved by factorization
X(x-2) - 2(x-2) =0
Hence;
(x-2) (x-2) =0
x= 2 (twice)
(2x-2)^2 = 4 can be solved by taking square root
√[(2x-2)^2] = √4
2x-2 = 2
x= ±2
x^2 + 6x - 11 = 0 can be solved by completing the square
(x + 3)^2 = 11 + 3^2
(x + 3)^2 = 20 taking square root of both sides
x + 3 = ±√20
x + 3 = ±2√5
x= -3 ±2√5
2x^2 - x + 5 = 0 can be solved by formula method
Using
x= -b±√b^2 - 4ac/2a
a= 2, b = -1, c= 5
x= 1 ± √1 - 40/4
x= 1 ± √-39/4
x= 1 ± 39i/4