If i have decent grades, but miss a lot of school will they fail me???

Mathematics

1 answer:

BaLLatris [955]2 years ago

4 0

Answer:

I think it's no

Step-by-step explanation:

because you will study hard

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When flying at an altitude of 5 miles, the lines of sight to the horizon looking north and south make about a 173.7 degree angle

oksian1 [2.3K]

5 miles high is one of the sides of a triangle depending on accuracy level
h^2=x^2+y^2
we don't have 2 distances
Tan A=O/a
O=a tan A
We solve for O because the angle is at the top of the line going up and we want the opposite angle that is along the ground
O=5×tan(173.7/2)=90.854033512
The distance he can see is:
90.85*2~181.7 miles

Now we need to find the distance between lines:
The north south distance between each line is 69 miles
thus the number of degrees he will see will be:
181.7/69
=2 19/30

7 0

3 years ago

Select interior, exterior, or on the circle (x - 5) 2 + (y + 3) 2 = 25 for the following point. (2, 3)

lozanna [386]

Standard equation of a circle: (x-h)² + (y-k)² = r² where (h, k) is the center and r is the radius. In the case of our equation here, (x-5)² + (y+3)² = 25, we can conclude that our circle has a center at (5, -3) and a radius of 5 units.

We can use the distance formula with the center (5, -3) and our point (2, 3) to see how far away they are...if the distance between them is less than the radius of the circle, it is on the interior. If it's equal, it's on the circle. If it's greater, it's on the exterior.

Distance = $\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Distance = $\sqrt{(-3-3)^2+(5-2)^2}$

Distance = $\sqrt{(-6)^2+3^2}$

Distance = $\sqrt{36+9}$

Distance = $\sqrt{45}\approx6.7082$

$6.7082>5,\ so\ (2,3)\ is\ on\ the\ \boxed{exterior}$