Given an equation x3 - 8x2 - 9x + 72 = 0, find the zeros.
1 answer:
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Answer:
b.
Step-by-step explanation:
Here we are given and are asked to simplify it. We are going to apply the law of exponents which says that
Hence applying this rule in our problem we get, where a=y and n=3
Hence our C) Option is correct
Problem 1
Since point P is the tangent point, this means angle OPT is a right angle
angle OPT = 90 degrees
Let's use the Pythagorean theorem to find the missing side 'a'
a^2 + b^2 = c^2
a^2 + 7^2 = 12^2
a^2 + 49 = 144
a^2 = 144 - 49
a^2 = 95
a = sqrt(95)
a = 9.7467943
a = 9.7
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Let's use the sine and arcsine rule to find angle y
sin(angle) = opposite/hypotenuse
sin(y) = 7/12
y = arcsin( 7/12 )
y = 35.6853347
y = 35.7
The value of y is approximate. Make sure your calculator is in degree mode. Arcsine is the same as inverse sine or
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Problem 2
Focus on triangle OHP. This may or may not be a right triangle. The goal is to test if it is or not.
We use the converse of the Pythagorean theorem to check.
Recall that the converse of the Pythagorean theorem says: If a^2+b^2 = c^2 is a true equation, then the triangle is a right triangle. The value of c is always the longest side. The order of a and b doesn't matter.
In this case we have
- a = 7
- b = 13
- c = 16
Which leads to
a^2 + b^2 = c^2
7^2 + 13^2 = 16^2
49 + 169 = 256
218 = 256
We get a false equation at the end, since both sides aren't the same number, which means the original equation is false.
We don't get a^2 + b^2 = c^2 to be true, therefore this triangle is not a right triangle.
Consequently, this means angle OPH cannot possible be 90 degrees (if it was then we'd have a right triangle). Therefore, point P is not a tangent point.
You follow the same basic idea for triangle OQH and show that point Q is not a tangent point either. Or you could use a symmetry argument to note that triangle OPH is a mirror reflection of triangle OQH over the line segment OH. This implies that whatever properties triangle OPH has, then triangle OQH has them as well for the corresponding pieces.
Treat 
as the boundary of the region 
, where is the part of the surface 
bounded by . We write
with
.
By Stoke's theorem, the line integral is equivalent to the surface integral over of the curl of 
. We have
so the line integral is equivalent to
where
is a vector-valued function that parameterizes . In this case, we can take
with
and 
. Then
and the integral becomes
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with
By Stoke's theorem, the line integral is equivalent to the surface integral over
so the line integral is equivalent to
where
with
and the integral becomes