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zimovet [89]
3 years ago
9

A ship at position (1, 0) on a nautical chart (with north in the positive y direction) sights a rock at position (7, 5). What an

gle θ does this vector make with due north? (This is called the bearing of the rock from the ship. Round your answer to two decimal places.) answer should be in radians east of north
Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
3 0

Answer:

0.69473828 radians east of north.

Step-by-step explanation:

Interpreting the information given on a Cartesian plane, the position of the ship will be (1,0) and that of the rock will be (7,5). Connecting the point (1,0) and (7,5) with a line and dropping a perpendicular line from point (7,5) to the x-axis to intersect the x-axis at (7,0), will form a right-triangle with base length 6 units and a height of 5 units.

<u>Finding the base length from points (1,0) and (7,0)</u>

d=√(x₂-x₁)²+(y₂-y₁)²

d=√(7-1)²

d= √6²

d=6 units

<u>Finding the height from points (7,5) and (7,0)</u>

d=√(x₂-x₁)²+(y₂-y₁)²

d=√(7-7)² + (0-5)²

d=√-5²

d=√25

d=5 units

So you now have a right triangle with base length 6 units and height 5 units.

To get the bearing of rock from ship position you apply the formula for tangent of an angle

Tan Ф = length of opposite side/length of adjustment side

TanФ=5/6

Tan Ф =0.83333333333

Tan⁻¹ (0.83333333333) =0.69473828 radians east of north.

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Express each answer in simplified radical form √24+4√54=?
VMariaS [17]

Answer:

The answer to your question is:    14\sqrt{6}

Step-by-step explanation:

                                            √24+4√54

                            \sqrt{2^{2}(2)(3)} + 4\sqrt{3^{2}(3)(2) }

                                     2\sqrt{6}  + 12\sqrt{6}

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3 years ago
R (-3,1) and S (-1,3) are points on a circle. If RS is a diameter, find the equation of the circle.​
ELEN [110]

Answer:

\sf (x+2)^2+(y-2)^2=2

Step-by-step explanation:

If RS is the diameter of the circle, then the midpoint of RS will be the center of the circle.

\sf midpoint=\left(\dfrac{x_s-x_r}{2}+x_r,\dfrac{y_s-y_r}{2}+y_r \right)

             \sf =\left(\dfrac{-1-(-3)}{2}+(-3),\dfrac{3-1}{2}+1 \right)

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Equation of a circle:   \sf (x-h)^2+(y-k)^2=r^2

(where (h, k) is the center and r is the radius)

Substituting found center (-2, 2) into the equation of a circle:

\sf \implies (x-(-2))^2+(y-2)^2=r^2

\sf \implies (x+2)^2+(y-2)^2=r^2

To find \sf r^2, simply substitute one of the points into the equation and solve:

\sf \implies (-3+2)^2+(1-2)^2=r^2

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\sf (x+2)^2+(y-2)^2=2

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2 years ago
Someone help me with this still im stock on it plz
pav-90 [236]

Answer:

a>4 or a < 8

Step-by-step explanation:

We can start with the inequality on the left

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subtract 8 from both sides to isolate the variable and its coefficient

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divide both sides by 5 to isolate a

a > 4

moving on to the inequality on the right

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treat the minus sign as a -1

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add 4 to both sides to isolate a and its coefficient

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multiply both sides by -1 but change the sign because we are multiplying by a negative number

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Equation: In algebra, an equation can be defined as a mathematical statement consisting of an equal symbol between two algebraic expressions that have the same value. The most basic and common algebraic equations in math consist of one or more variables.


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