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Answer: the angle is 35 degrees
The supplementary angle is 145 degrees
Step-by-step explanation:
Supplementary angles are angles whose sum is 180 degrees. This means that if angle x and angle y are supplementary, then the sum of angle x + angle y is 180 degrees.
The given angle measures 110° less than the measure of its supplementary angle.
Let the angle be x degrees and the supplementary angle be y degrees. This means that.
x = y - 110
Recall, sum of supplementary angles is 180 degrees. Therefore,
x + y = 180 - - - - - - - - 1
Substituting x = y - 110 into equation 1, it becomes
y - 110 + y = 180
2y -110 = 180
2y = 180 + 110 = 290
y = 290/2 = 145 degrees
x = y - 110 = 145 - 110
x = 35 degrees
<span>place vertex at (0,0)
basic equation: x^2=4py
p=distance from vertex to focus)
y=at ends of parabolic cross section
y = x^2/4p
Given y= 1/48x^2
So,
1/48x^2=X*2/4p
1/4p = 1/48
4p = 48
p = 48/4 = 12 ft</span>
Answer: the rate at which its altitude is changing is -47 ft/s
Step-by-step explanation:
Given that;
ds/dt = 45 mph = 45 × 5280/3600 = 66 ft/sec
no we have to determine dx/dt
so
cos∅ = dx/dt / ds/dt
cos45° = dx/dt / 66
dx/dt = 0.7071 × 66
dx/dt = 46.6686 ≈ 47 ft/s ( as altitude is decreasing, its negative - )
so the rate at which its altitude is changing is -47 ft/s
Assuming the question is asking about $y=2+(x-3)^2$
Answer:
$(3, \infty)$
Step-by-step explanation:
We see that the function is quadratic, and it is already in vertex form. This is good, as it saves a step. Because anything squared is always positive or $0$, when we minimize this function we want to have the thing inside the square be $0$. This means $x-3=0$, so $x=3$.
We can plug in for $x$ to find the vertex.
$y=2+(3-3)^2$
$y=2+0^2$
$y=2+0$
$y=2$.
This means that $(3, 2)$ is the vertex of the parabola(a quadratic equation in $x$).
We see that the coefficient of the highest order term(the $x^2$) will be positive, so the parabola faces upward. This means it will always be increasing to the right of the vertex, and decreasing to the left of the vertex.
So, the interval where $y=2+(x-3)^2$ is increasing is the interval to the right of the vertex, which is $(3, \infty)$