In this exercise, we need to find the angle ∠EDH, that is, according to the figure above, this angle is:
∠EDH = (5x)°
So, our goal is to find the value of x. By taking a look in the figure above, this equation is true:
(5x)° + (4x)° = 90°
Therefore:
(9x)° = 90° ∴ x = 10
So:
∠EDH = (5x)° ∴ ∠EDH = [(5)(10)]° ∴ ∠EDH = 50°
Finally, the right answer is 50°
One way to find the slope-intercept form is to plug the given values into point-slope form, y - y_{1} = m (x - x_{1}), where x_{1} and y_{1} are the coordinate points and m is the slope. Then, you solve for y.
y - y_{1} = m (x - x_{1}) Plug in the values
y - (-4) = 34 (x - 5) Fix the minus negative four
y + 4 = 34 (x - 5) Use the Distributive Property
y + 4 = 34x - 170 Subtract 4 from both sides
y = 34x - 174
The slope-intercept form of the equation that passes through (5, -4) and has a slope of 34 is y = 34x - 174.
Answer:
Cos x
Step-by-step explanation:
Answer: the rate at which the distance between the boats is increasing is 68 mph
Step-by-step explanation:
The direction of movement of both boats forms a right angle triangle. The distance travelled due south and due east by both boats represents the legs of the triangle. Their distance apart after t hours represents the hypotenuse of the right angle triangle.
Let x represent the length the shorter leg(south) of the right angle triangle.
Let y represent the length the longer leg(east) of the right angle triangle.
Let z represent the hypotenuse.
Applying Pythagoras theorem
Hypotenuse² = opposite side² + adjacent side²
Therefore
z² = x² + y²
To determine the rate at which the distances are changing, we would differentiate with respect to t. It becomes
2zdz/dt = 2xdx/dt + 2ydy/dt- - - -- - -1
One travels south at 32 mi/h and the other travels east at 60 mi/h. It means that
dx/dt = 32
dy/dt = 60
Distance = speed × time
Since t = 0.5 hour, then
x = 32 × 0.5 = 16 miles
y = 60 × 0.5 = 30 miles
z² = 16² + 30² = 256 + 900
z = √1156
z = 34 miles
Substituting these values into equation 1, it becomes
2 × 34 × dz/dt = (2 × 16 × 32) + 2 × 30 × 60
68dz/dt = 1024 + 3600
68dz/dt = 4624
dz/dt = 4624/68
dz/dt = 68 mph
Answer:
let A (2, 6) be (x1, y1)
B (-5, 2) be (x2, y2)
distance AB =√[(x2 - x1)² + (y2 - y1)²]
√[(-7)²+ (-4)²]
√65
