Answer:
<h2>2.2</h2>
Step-by-step explanation:
Use the cosine law:
We have:
Substitute:
Answer:
2
Step-by-step explanation:
2^3= 2*2*2
factors of 8= 1,2,4,8
factors of 2= 1,2
Answer:
Parallel - 8
Perpendicular - -1/8
Step-by-step explanation:
If a line is parallel to another then they have the same slope
If a line is perpendicular to another line then the slope is just flipped and so is the sign
25/3 ft/s is speed of the tip of his shadow moving when a man is 40 ft from the pole given that a street light is mounted at the top of a 15-ft-tall pole and the man is 6 ft tall who is walking away from the pole with a speed of 5 ft/s along a straight path. This can be obtained by considering this as a right angled triangle.
<h3>How fast is the tip of his shadow moving?</h3>
Let x be the length between man and the pole, y be the distance between the tip of the shadow and the pole.
Then y - x will be the length between the man and the tip of the shadow.
Since two triangles are similar, we can write
⇒15(y-x) = 6y
15 y - 15 x = 6y
9y = 15x
y = 15/9 x
y = 5/3 x
Differentiate both sides
dy/dt = 5/3 dx/dt
dy/dt is the speed of the tip of the shadow, dx/dt is the speed of the man.
Given that dx/dt = 5 ft/s
Thus dy/dt = (5/3)×5 ft/s
dy/dt = 25/3 ft/s
Hence 25/3 ft/s is speed of the tip of his shadow moving when a man is 40 ft from the pole given that a street light is mounted at the top of a 15-ft-tall pole and the man is 6 ft tall who is walking away from the pole with a speed of 5 ft/s along a straight path.
Learn more about similar triangles here:
brainly.com/question/8691470
#SPJ4
Let the lengths of the east and west sides be x and the lengths of the north and south sides be y. the dimensions you want are therefore x and y.
The cost of the east and west fencing is $4*2*x; the cost of the north and south fencing is $2*2*y. We have to put in that "2" because there are 2 sides that run from east to west and 2 sides that run from north to south.
The total cost of all this fencing is $4(2)(x) + $2(2)(y) = $128. Let's reduce this by dividing all three terms by 4: 2x + y = 32.
Now we are to maximize the area of the vegetable patch, subject to the constraint that 2x + y = 32. The formula for area is A = L * W. Solving 2x + y = 32 for y, we get y = -2x + 32.
We can now eliminate y. The area of the patch is (x)(-2x+32) = A. We want to maximize A.
If you're in algebra, find the x-coordinate of the vertex of this quadratic equation. Remember the formula x = -b/(2a)? Once you have calculated this x, subst. your value into the formula for y: y= -2x + 32.
Now multiply together your x and y values to obtain the max area of the patch.
If you're in calculus, differentiate A = x(-2x+32) with respect to x and set the derivative equal to zero. This approach should give you the same x value as before; the corresponding y value will be the same; y=-2x+32.
Multiply x and y together. That'll give you the maximum possible area of the garden patch.
