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2 answers:
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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.
Answer:
C. 15
Step-by-step explanation:
The following proportional can be set-up for the figure given:
Solve for x
Cross multiply
15 = 6x - 5x
15 = x
x = 15
9514 1404 393
Answer:
possible rational roots: ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12}
actual roots: -1, (2 ±4i√2)/3
no turning points; no local extrema
end behavior is same-sign as x-value end-behavior
Step-by-step explanation:
The Fundamental Theorem tells us this 3rd-degree polynomial will have 3 roots.
The Rational Root Theorem tells us any rational roots will be of the form ...
±{factor of 12}/{factor of 3} = ±{1, 2, 3, 4, 6, 12}/{1, 3}
= ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12} . . . possible rational roots
Descartes' Rule of Signs tells us the two sign changes mean there will be 0 or 2 positive real roots. Changing signs on the odd-degree terms makes the sign-change count go to 1, so we know there is one negative real root.
The y-intercept is 12. The sum of all coefficients is 22, so f(1) > f(0) and there are no positive real roots in the interval [0, 1]. Synthetic division by x-1 shows the remainder is 22 (which we knew) and all the quotient coefficients are all positive. This means x=0 is an upper bound on the real roots.
The sum of odd-degree coefficients is 3+8=11, equal to the sum of even-degree coefficients, -1+12=11. This means that -1 is a real root. Synthetic division by x+1 shows the remainder is zero (which we knew) and the quotient coefficients alternate signs. This means x=-1 is a lower bound on real roots. The quotient of 3x^2 -4x +12 is a quadratic factor of f(x):
f(x) = (x +1)(3x^2 -4x +12)
The complex roots of the quadratic can be found using the quadratic formula:
x = (-(-4) ±√((-4)^2 -4(3)(12)))/(2(3)) = (4 ± √-128)/6
x = (2 ± 4i√2)/3 . . . . complex roots
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The graph in the third attachment (red) shows there are no turning points, hence no relative extrema. The end behavior, as for any odd-degree polynomial with a positive leading coefficient, is down to the left and up to the right.
