the bearing of c from a is 210°
(i) find the bearing of B from A
(ii) find the bearing of A from B
if you could also solve the question above I'd be very grateful
I need it in at least 10 minutes so please answer it!! ♡♡
Answers
Answer:
(a) His speed when he runs from C to A is [tex]4.\overline {285714}[/tex] m/s
(i) The bearing of B from A is approximately 127.18°
(ii) The bearing of A from B is approximately 307.18°
Step-by-step explanation:
(a) The given parameters are;
The distance from A to B = 120 m
The speed with which Olay runs from A to B, v₁ = 4 m/s
The distance from B to C = 180 m
The speed with which Olay runs from B to C, v₂ = 3 m/s
The distance from C to A = 150 m
His average speed for the whole journey = 3.6 m/s
We find
The total distance of running from A back to A, d = 120 m + 180 m + 150 m = 450 m
The time it takes to run from A to B, t₁ = 120 m/(4m/s) = 30 seconds
The time it takes to run from B to C, t₂ = 180 m/(3m/s) = 60 seconds
Let t₃ represent the time it takes Olay to run from C to A
We have;
The total time it takes to run from A back to A = t₁ + t₂ + t₃
Therefore;
[tex]Average \ velocity = \dfrac{Total \ distance }{Total \ time} = \dfrac{d}{t_1 + t_2 + t_3}[/tex]
Substituting the known values for the average velocity, 'd', 't₁' and 't₂' gives;
[tex]Average \ velocity = 3.6 \, m/s = \dfrac{450 \, m}{30 \, s + 60 \, s + t_3}[/tex]
3.6 m/s × (30 s + 60 s + t₃) = 450 m
3.6 m/s × 30 s + 3.6 m/s × 60 s + 3.6 m/s × t₃ = 450 m
108 m + 216 m + 3.6 m/s × t₃ = 450 m
∴ 3.6 m/s × t₃ = 450 m - (108 m + 216 m) = 126 m
t₃ = 126 m/(3.6 m/s) = 35 s
The speed with which Olay runs from C to A, v₃ = Distance from C to A/t₃
The speed with which he runs from C to A = 150 m/(35 s) = 30/7 m/s =[tex]4.\overline {285714}[/tex] m/s
(i) The given bearing of C from A = 210°
By cosine rule, we have;
a² = b² + c² - 2·b·c·cos(A)
∴ cos(A) = (b² + c² - a²)/(2·b·c)
Where;
a = The distance from B to C = 180 m
b = The distance from C to A = 150 m
c = The distance from A to B = 120 m
We find;
cos(A) = (150² + 120² - 180²)/(2 × 150 × 120) = 0.125
A = arccos(0.125) ≈ 82.82°
The bearing of B from A ≈ 210° - 82.82° ≈ 127.18°
The bearing of B from A ≈ 127.18°
(ii) The angle, θ, formed by the path of the bearing of A from B is an alternate to the supplementary angle of the bearing of B from A
Therefore, we have;
θ ≈ 180°- 127.18° ≈ 52.82°
The bearing of A from B = The sum of angle at a point less θ
∴ The bearing of A from B = 360° - 52.82° ≈ 307.18°
The bearing of A from B ≈ 307.18°.
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A line segment has endpoints S(-9,-4) and T(6,5). Point R lies on ST such that the ratio of SR to RT is 2:1. What are the coordinates of point R?
Answers
Answer:
Coordinates of R = (1,2)
Step-by-step explanation:
Let the coordinate of R be (x, y)
Since coordinates of S and T are S(-9,-4) and T(6,5). Then we can use the Formula for length of line from coordinates to find coordinates of R since SR/RT = 2/1
Thus; 1(S) = 2(T)
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Coordinates of R = (3/3), (6/3)
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Step-by-step explanation:
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Step-by-step explanation:
The first term in an arithmetic sequence is 5. The fourth term in the sequence is −4. The tenth term
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Which function can be used to find the nth term of the arithmetic sequence?
Answers
Answer:
aₙ = -3n + 8Step-by-step explanation:
The nth term of an arithmetic sequence: aₙ = a₁ + d(n - 1)
{d = common difference}
a₁ = 5
a₄ = 5 + d(4 - 1)
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-4 - 5 = 3d
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d = -3
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aₙ = 5 + (-3)(n - 1)
aₙ = -3n + 8
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How to determine the function of the nth term?The given parameters are:
a = 5, first term
T4 = -4 --- the 4th term
T10 = -22 --- the 10th term
The nth term of an arithmetic sequence is:
Tn = a + (n - 1) * d
So, we have:
T4 = a + (4 - 1) * d
Substitute known values
-4 = 5 +(4 -1) * d
This gives
-4 = 5 + 3d
Subtract 5 from both sides
3d = -9
Divide by 3
d = -3
Recall that:
Tn = a + (n - 1) * d
So, we have:
Tn = 5 + (n - 1) * -3
Expand
Tn = 5 + 3 - 3n
Solve
Tn = 8 - 3n
Hence, the nth term of the sequence is Tn = 8 - 3n
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Subtract
20
from both sides of the equation.
x
2
−
4
x
=
−
20
To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of
b
.
(
b
2
)
2
=
(
−
2
)
2
Add the term to each side of the equation.
x
2
−
4
x
+
(
−
2
)
2
=
−
20
+
(
−
2
)
2
Simplify the equation.
Tap for more steps...
x
2
−
4
x
+
4
=
−
16
Factor the perfect trinomial square into
(
x
−
2
)
2
.
(
x
−
2
)
2
=
−
16
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x
.
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x
=
2
±
4
i
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Answers
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Answer from jmonterrozar
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Answers
Answer:
2879ft
Step-by-step explanation:
The level runway (x), height gained (h), and distance travelled (r) form a right-angled triangle with base angle
10
∘
.
We may hence use trig ratios to solve for any of the unknowns, in particular for the distance r as follows :
Find the area of the trapezoids
Answers
Answer:
48 square units
Step-by-step explanation:
Each square on the grid is 2 units by 2 units.
Bases: AT = 8; RS = 4
Height: 8
area = (b1 + b2)h/2
area = (8 + 4)(8)/2
area = 48 square units
Answer:
The area of trapezoid is 48 sq units.
Step-by-step explanation:
According to the question,
➻ Trapezoid is consist of many square shapes. Each grid square equals 2 units. We need to find the area of trapezoid.
➻ To find area we need to find the length if base and height of trapezoid.
Calculate the square grids to find base and height. we get.
Base ; SR = 4 units and TA = 8units
Height = 8 units.
Finding the area of trapezoid.
We know that,
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plug the values of both base and a height.
Area of trapezoid = ( 4 + 8 ) × 8 / 2
Add the numbers which are in bracket and multiply by 8.
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Divide 96 by 2.
Area of trapezoid = 48 sq units.
Hence,
The area of trapezoid is 48 sq units.