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Inessa05 [86]
2 years ago
5

A man knows that his apple tree is 51 ft high. At 11:00 am the tree casts a shadow of 68 ft. What is the angle of elevation of t

he sun? Round to the nearest tenth
Mathematics
1 answer:
frozen [14]2 years ago
6 0

Answer:

36.9 degrees

Step-by-step explanation:

This problem can be solved by drawing a right triangle and doing trigonometry. We can draw our triangle with a height of 51 and a base of 68. In order to find the angle from the ground to the sun we can use arctangent (tan^-1). Since tan(angle)=opposite/adjacent we can find the angle by doing angle=tan^-1(51/68)

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AXY Zand AABCare similar triangles. Given the dimensions shown in the diagram, what is the
Lady bird [3.3K]

Answer:

3.528

Step-by-step explanation:

First, you need to determine the scale factor. Divide one side in the larger triangle by the corresponding side in the smaller triangle. For example, 8/2. This would give you a scale factor of 2. Then you can reverse the process with the dimensions you need to get the area. The height of the larger triangle is 3.36, so to find the height of the smaller one, you divide it by 2, or the scale factor, to get the height of the smaller triangle which is 1.68. Next, you can solve for area of triangle using a=bh/2, so 4.2*1.68/2=a = 3.528.

6 0
3 years ago
Halp pls ok<br><br> x =x=x, equals <br> ^\circ <br> ∘
Ber [7]

Answer:

60

Step-by-step explanation:

The yellow and green angles form a right angle, so the measures of those combined will be 90 degrees.


90 - 30 = 60

7 0
2 years ago
Let v = (v1, v2) be a vector in R2. Show that (v2, −v1) is orthogonal to v, and use this fact to find two unit vectors orthogona
andrey2020 [161]

Answer:

a. v.v' = v₁v₂ -  v₁v₂ = 0 b.  (20, -21)/29 and  (-20,21)/29

Step-by-step explanation:

a. For two vectors a, b to be orthogonal, their dot product is zero. That is a.b = 0.

Given v = (v₁, v₂) = v₁i + v₂j and v' =  (v₂, -v₁) = v₂i - v₁j, we need to show that v.v' = 0

So, v.v' = (v₁i + v₂j).(v₂i - v₁j)

= v₁i.v₂i + v₁i.(- v₁j) + v₂j.v₂i + v₂j.(- v₁j)

= v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j

i.i = 1, i.j = 0, j.i = 0 and j.j = 1

So, v.v' = v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j  

= v₁v₂ × 1 - v₁v₁ × 0 + v₂v₂ × 0 - v₂v₁ × 1

= v₁v₂ - v₂v₁

=  v₁v₂ -  v₁v₂ = 0

So, v.v' = 0

b. Now a vector orthogonal to the vector v = (21,20) is v' = (20,-21).

So the first unit vector is thus a = v'/║v'║ = (20, -21)/√[20² + (-21)²] = (20, -21)/√[400 + 441] = (20, -21)/√841 = (20, -21)/29.

A unit vector perpendicular to a and parallel to v is b = (-21, -20)/29. Another unit vector perpendicular to b, parallel to a and perpendicular to v is thus a' = (-20,-(-21))/29 = (-20,21)/29

8 0
2 years ago
PLEASE PLEASE PLEASE HELP ME!!!!! PLEASE PLEASE ILL GIVE POINTS, BRAINLY, AND THANKS
Murljashka [212]

Composing function is applying one function to the result of another and an arithmetic operation is when you add, subtract, multiply, or divide two functions.

8 0
2 years ago
How do you determine if a relation is quadratic by calculation differences, analyzing a graph, examining the degree in an equati
Kryger [21]

1) We can determine by the table of values whether a function is a quadratic one by considering this example:

x | y 1st difference 2nd difference

0 0 3 -0 = 3 7-3 = 4

1 3 10 -3 = 7 11 -7 = 4

2 10 21 -10 =11 15 -11 = 4

3 21 36-21 = 15 19-5 = 4

4 36 55-36= 19

5 55

2) Let's subtract the values of y this way:

3 -0 = 3

10 -3 = 7

21 -10 = 11

36 -21 = 15

55 -36 = 19

Now let's subtract the differences we've just found:

7 -3 = 4

11-7 = 4

15-11 = 4

19-15 = 4

So, if the "second difference" is constant (same result) then it means we have a quadratic function just by analyzing the table.

3) Hence, we can determine if this is a quadratic relation calculating the second difference of the y-values if the second difference yields the same value. The graph must be a parabola and the highest coefficient must be 2

4 0
1 year ago
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