Write the given equation in standard form. −7x−5y=3
1 answer:
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<h3>Answer: 24/25</h3>
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Explanation:
Sine is given to be negative, and so is tangent. This only happens in quadrant Q4
Recall that y = sin(theta), so if sin(theta) < 0, then we're below the x axis.
If tan(theta) < 0, then this means cos(theta) > 0
So we have y < 0 and x > 0 which places the angle somewhere in Q4.
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Draw a right triangle as shown below in the attached image. We have AC = 25 and BC = 7. Use the pythagorean theorem to find that AB = 24
So this is what your steps may look like
a^2+b^2 = c^2
7^2+b^2 = 25^2
b^2+49 = 625
b^2 = 625-49
b^2 = 576
b = sqrt(576)
b = 24
So because AB = 24, we know that the cosine of the angle is adjacent/hypotenuse = 24/25
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As an alternative, you could use the trig identity
sin^2(x) + cos^2(x) = 1
and plug in the given value of sine to solve for cosine. The cosine value result will be positive since we're in Q4.
So,
sin^2(x) + cos^2(x) = 1
(-7/25)^2 + cos^2(x) = 1
(49/625) + cos^2(x) = 1
cos^2(x) = 1 - (49/625)
cos^2(x) = (625/625) - (49/625)
cos^2(x) = (625-49)/625
cos^2(x) = 576/625
cos(x) = sqrt(576/625)
cos(x) = sqrt(576)/sqrt(625)
cos(x) = 24/25
This is effectively a rephrasing of the previous section since the pythagorean trig identity is more or less the pythagorean theorem (just in a trig form)
The given quadratic describes a parabola that opens upward. Its one absolute extreme is a minimum that is found at x = -3/2. The value of the function there is
(-3/2 +3)(-3/2) -1 = -13/4
The one relative extreme is a minimum at (-1.5, -3.25).
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For the parabola described by ax² +bx +c, the vertex (extreme) is found where
x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.
(-3/2 +3)(-3/2) -1 = -13/4
The one relative extreme is a minimum at (-1.5, -3.25).
_____
For the parabola described by ax² +bx +c, the vertex (extreme) is found where
x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.