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3 years ago

Aiko stands 5 m from a tree. At that distance, the angle of elevation from the ground to the top of the tree is 80°.

Mathematics

1 answer:

denis23 [38]3 years ago

3 0

Answer:

$\boxed {\boxed {\sf About \ 28.36 \ meters }}$

Step-by-step explanation:

Assuming the tree is perpendicular to the ground, we can use the right triangle trigonometric ratios to find the tree's height.

sin(θ)= opposite/hypotenuse
cos(θ)= adjacent/hypotenuse
tan(θ)= opposite/adjacent

Now, let's draw a diagram. We know Aiko is 5 meters from the base of the tree. From there, the angle to the top of the tree is 80 degrees. We are looking for x, the tree's height. The diagram attached is not to scale.

We base the sides off of the angle. x is opposite of 80 degrees and 5 is adjacent. Therefore we must use tangent.

$tan(\Theta)=\frac{opposite}{adjacent}$

opposite=x
adjacent=5 m
θ=80

Substitute in the known variables.

$tan(80)=\frac{x}{5 \ m }$

We want to find x, the height of the tree. Therefore we need to isolate that variable.

x is being divided and the inverse operation is multiplication. Multiply both sides of the equation by 5 meters.

$5 \ m* tan(80)=\frac{x}{5 \ m }* 5 \ m$

$5 \ m* tan(80)=x$

$5 \ m *5.67128182=x$

$28.3564091 \ m = x$

The question asks for an approximation, so let's round to the nearest hundredth.

The 6 in the thousandth place tells us to round the 5 to a 6.

$28.36 \ m \approx x$

The tree is about <u>28.36 meters tall.</u>

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3 years ago

The Greenpoint factory produced two-fifths of the Consolidated Brick Company’s bricks in 1991. If the Greenpoint factory produce

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Answer:

3500 tons

Step-by-step explanation:

The Greenpoint factory produced 2/5 of the bricks that the Consolidate Brick Company produced in 1991.

Let the amount of bricks produced by the Greenpoint factory be g and the amount of bricks produced by the Consolidated brick company be c.

Therefore:

g = 2/5 * c = 2c/5

That year, the Greenpoint factory produced 1400 tons of bricks. This implies that:

1400 = 2c/5

To find the amount that the Consolidated Brick Company produced, solve for c:

1400 = 2c/5

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3 years ago

I have an assignment and I am having trouble with it. Can someone please help ASAP???

bezimeni [28]

Answer:

A) Find the sketch in attachment.

In the sketch, we have plotted:

- The length of the arena on the x-axis (90 feet)

- The width of the arena on the y-axis (95 feet)

- The position of the robot at t = 2 sec (10,30) and its position at t = 8 sec (40,75)

The origin (0,0) is the southweast corner of the arena. The system of inequalities to descibe the region of the arena is:

$0\leq x \leq 90\\0\leq y \leq 95$

Since the speed of the robot is constant, it covers equal distances (both in the x- and y- axis) in the same time.

Let's look at the x-axis: the robot has covered 10 ft in 2 s and 40 ft in 8 s. There is a direct proportionality between the two variables, x and t:

$\frac{10}{2}=\frac{40}{8}$

So, this means that at t = 0, the value of x is zero as well.

Also, we notice that the value of y increases by $\frac{75-30}{8-2}=7.5 ft/s$ (7.5 feet every second), so the initial value of y at t = 0 is:

$y(t=0)=30-7.5\cdot 2 =15 ft$

So, the initial position of the robot was (0,15) (15 feet above the southwest corner)

The speed of the robot is given by

$v=\frac{d}{t}$

where d is the distance covered in the time interval t.

The distance covered is the one between the two points (10,30) and (40,75), so it is

$d=\sqrt{(40-10)^2+(75-30)^2}=54 ft$

While the time elapsed is

$t=8 sec-2 sec = 6 s$

Therefore the speed is

$v=\frac{54}{6}=9 ft/s$

The equation for the line of the robot is:

$y=mx+q$

where m is the slope and q is the y-intercept.

The slope of the line is given by:

$m=\frac{75-30}{40-10}=1.5$

Which means that we can write an equation for the line as

$y=mx+q\\y=1.5x+q$

where q is the y-intercept. Substituting the point (10,30), we find the value of q:

$q=y-1.5x=30-1.5\cdot 10=15$

So, the equation of the line is

$y=1.5x+15$

By prolonging the line above (40,75), we see that the line will hit the north wall. The point at which this happens is the intersection between the lines

$y=1.5x+15$

and the north wall, which has equation

$y=95$

By equating the two lines, we find:

$1.5x+15=95\\1.5x=80\\x=\frac{80}{15}=53.3 ft$

So the coordinates of impact are (53.3, 95).

The distance covered between the time of impact and the initial moment is the distance between the two points, so:

$d=\sqrt{(53.5-0)^2+(95-15)^2}=95.7 ft$

From part B), we said that the y-coordinate of the robot increases by 15 feet/second.

We also know that the y-position at t = 0 is 15 feet.

This means that the y-position at time t is given by equation:

$y(t)=15+7.5t$

The time of impact is the time t for which

y = 95 ft

Substituting into the equation and solving for t, we find:

$95=15+7.5t\\7.5t=80\\t=10.7 s$

The path followed by the robot is sketched in the second graph.

As the robot hits the north wall (at the point (53.3,95), as calculated previously), then it continues perpendicular to the wall, this means along a direction parallel to the y-axis until it hits the south wall.

As we can see from the sketch, the x-coordinate has not changed (53,3), while the y-coordinate is now zero: so, the robot hits the south wall at the point

(53.3, 0)

The perimeter of the triangle is given by the sum of the length of the three sides.

- The length of 1st side was calculated in part F: $d_1 = 95.7 ft$

- The length of the 2nd side is equal to the width of the arena: $d_2=95 ft$

- The length of the 3rd side is the distance between the points (0,15) and (53.3,0):

$d_3=\sqrt{(0-53.3)^2+(15-0)^2}=55.4 ft$

So the perimeter is

$d=d_1+d_2+d_3=95.7+95+55.4=246.1 ft$

The area of the triangle is given by:

$A=\frac{1}{2}bh$

where:

$b=53.5 ft$ is the base (the distance between the origin (0,0) and the point (53.3,0)

$h=95 ft$ is the height (the length of the 2nd side)

Therefore, the area is:

$A=\frac{1}{2}(53.5)(95)=2541.3 ft^2$

The percentage of balls lying within the area of the triangle traced by the robot is proportional to the fraction of the area of the triangle with respect to the total area of the arena, so it is given by:

$p=\frac{A}{A'}\cdot 100$

where:

$A=2541.3 ft^2$ is the area of the triangle

$A'=90\cdot 95 =8550 ft^2$ is the total area of the arena

Therefore substituting, we find:

$p=\frac{2541.3}{8550}\cdot 100 =29.7\%$

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3 years ago

A steel company sells 48 tons of steel each year.

ddd [48]

Answer:

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Step-by-step explanation:

To solve the first step is to change years to months because the time periods are in different units. There are 12 months in 1 year. This means that the company sells 48 tons over 12 months. 4 months is the equivalent of 1/3 of 12 months. This means that we can find the amount sold in 4 months by dividing 48 by 3. 48/3 is 16. Thus, the steel company sells 16 tons in 4 months.

Another way to find the answer is to find the unit rate. The unit rate is how many tons are sold in 1 month. Find this by dividing 48 by 12 because this represents the sales of 1 month. 48/12=4, this means that in 1 month, 4 tons of steel are sold. Next, multiply 4 times 4 to find the amount of steel sold in 4 months. 4 times 4 also gives you the same answer of 16.

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Ostrovityanka [42]

Answer:

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Step-by-step explanation:

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