
When variables with exponents are divided by each other, you subtract the exponents.
For example:


Cos(x) = sin(90 - x)
cos(53) = sin(90 - 53)
cos(53) = sin(37)
The equation for cosine is <span><span><span>cos<span>(x)</span></span>=<span>Adjacent/Hypotenuse
</span></span></span>The inside trig function is <span><span>arccos<span>(<span>3/5</span>)</span></span></span>, which means <span><span><span>cos<span>(x)</span></span>=<span>3/5</span></span></span>. Comparing <span><span><span>cos<span>(x)</span></span>=<span>Adjacent/Hypotenuse</span></span></span> with <span><span><span>cos<span>(x)</span></span>=<span>3/5
</span></span></span>
Find <span><span>Adjacent=3</span></span> and <span><span>Hypotenuse=5.
</span></span>Then, using the Pythagorean theorem, find <span><span>Opposite=?
</span></span>a² = c² - b²
a² = 5² - 3² = 25 - 9 = 16
a = √16 = 4
<span><span>Adjacent=3</span></span><span><span>Opposite=4</span></span><span><span>Hypotenuse=5
</span></span><span>
Plug in the value for sin(x) = opposite/hypotenuse
sin(x) = 4/5 </span>
Answer:
yes
Step-by-step explanation:
The line intersects each parabola in one point, so is tangent to both.
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For the first parabola, the point of intersection is ...
y^2 = 4(-y-1)
y^2 +4y +4 = 0
(y+2)^2 = 0
y = -2 . . . . . . . . one solution only
x = -(-2)-1 = 1
The point of intersection is (1, -2).
__
For the second parabola, the equation is the same, but with x and y interchanged:
x^2 = 4(-x-1)
(x +2)^2 = 0
x = -2, y = 1 . . . . . one point of intersection only
___
If the line is not parallel to the axis of symmetry, it is tangent if there is only one point of intersection. Here the line x+y+1=0 is tangent to both y^2=4x and x^2=4y.
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Another way to consider this is to look at the two parabolas as mirror images of each other across the line y=x. The given line is perpendicular to that line of reflection, so if it is tangent to one parabola, it is tangent to both.
Answer:
(0,-2), (5,0) and (10,2).
Step-by-step explanation:
Given equation is
.
Now we need to find 3 pairs of solutions in (x,y) form for the given equation.
As
is a linear equation so we are free to pick any number for x like x=0, 5, 10
Plug x=0 into
, we get:





Hence first solution is (0,-2)
We can repeat same process with x=5 and 10 to get the other solutions.
Hence final answer is (0,-2), (5,0) and (10,2).