If your Main Street line has a couple of points that you can find the coordinates ( x1, y1) and (x2, y2) you could determine the slope of the Main Street line : m= (y2-y1)/(x2-x1). Or maybe you can determine what the slope is from the line being a particular line , horizontal, etc. The slope of your path line would be m1 = -1/m. Use some point on this path to write the equation of a line in point- slope from. Suppose ( x3,y3) would be a point , ( look for example for the entrance coordinates), write y= m1(x-x3)+ y3 and plug in then values of m1, x3, y3 to have your path line equation. ~Nesabug :3
There would be 7 thousandths in the number 62.407 because the 7 is in the thousandths place.
Answer:
X =
Y = 204 =
Step-by-step explanation:
For question 4, ![sinA =\frac{a}{c}, cosA= \frac{b}{c}, tan B = \frac{b}{a}, sin J = \frac{j}{l}, cosK = \frac{j}{l}, tanK = \frac{k}{j}.](https://tex.z-dn.net/?f=sinA%20%3D%5Cfrac%7Ba%7D%7Bc%7D%2C%20cosA%3D%20%5Cfrac%7Bb%7D%7Bc%7D%2C%20tan%20B%20%3D%20%5Cfrac%7Bb%7D%7Ba%7D%2C%20sin%20J%20%3D%20%5Cfrac%7Bj%7D%7Bl%7D%2C%20cosK%20%3D%20%5Cfrac%7Bj%7D%7Bl%7D%2C%20tanK%20%3D%20%5Cfrac%7Bk%7D%7Bj%7D.)
Question 5. Option a and question 6. Option j
Step-by-step explanation:
Step 1:
The three basic formula needed to solve these questions are:
![sin\theta = \frac{oppositeside}{hypotenuse} , cos\theta = \frac{adjacentside}{hypotenuse}, tan\theta= \frac{opposite side}{adjacent side}.](https://tex.z-dn.net/?f=sin%5Ctheta%20%3D%20%5Cfrac%7Boppositeside%7D%7Bhypotenuse%7D%20%2C%20cos%5Ctheta%20%3D%20%5Cfrac%7Badjacentside%7D%7Bhypotenuse%7D%2C%20tan%5Ctheta%3D%20%5Cfrac%7Bopposite%20side%7D%7Badjacent%20side%7D.)
Step 2:
Using the above formula, we solve the following values
![=\frac{a}{c}.](https://tex.z-dn.net/?f=%3D%5Cfrac%7Ba%7D%7Bc%7D.)
![= \frac{b}{c}.](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bb%7D%7Bc%7D.)
![tanB= \frac{opposite side}{adjacent side}](https://tex.z-dn.net/?f=tanB%3D%20%5Cfrac%7Bopposite%20side%7D%7Badjacent%20side%7D)
![= \frac{b}{a}.](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bb%7D%7Ba%7D.)
![= \frac{j}{l}.](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bj%7D%7Bl%7D.)
![cosK = \frac{adjacentside}{hypotenuse}](https://tex.z-dn.net/?f=cosK%20%3D%20%5Cfrac%7Badjacentside%7D%7Bhypotenuse%7D)
![= \frac{j}{l}.](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bj%7D%7Bl%7D.)
![tanK= \frac{opposite side}{adjacent side}](https://tex.z-dn.net/?f=tanK%3D%20%5Cfrac%7Bopposite%20side%7D%7Badjacent%20side%7D)
![= \frac{k}{j}.](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bk%7D%7Bj%7D.)
Step 3:
For question 5, The triangle's angle = 23°, opposite side = BC inches and hypotenuse = 4 inches.
![sin\theta= \frac{opposite side}{hypotenuse}. sin 23^{\circ}= \frac{BC}{4}, sin23^{\circ} = 0.3907,BC = (0.3907)(4) = 1.5628.](https://tex.z-dn.net/?f=sin%5Ctheta%3D%20%5Cfrac%7Bopposite%20side%7D%7Bhypotenuse%7D.%20sin%2023%5E%7B%5Ccirc%7D%3D%20%5Cfrac%7BBC%7D%7B4%7D%2C%20sin23%5E%7B%5Ccirc%7D%20%3D%200.3907%2CBC%20%3D%20%280.3907%29%284%29%20%3D%201.5628.)
SO BC is 1.5628 inches, rounding this off to the nearest tenth, we get BC = 1.6 inches which is option a.
Step 4:
For question 6, The triangle's angle = 50°, opposite side = QR m the adjacent side = 8.1 m.
![tan\theta= \frac{opposite side}{adjacentside}. tan 50^{\circ}=\frac{QR}{8.1}, tan50^{\circ} = 1.1917,QR = (1.1917)(8.1) = 9.65277](https://tex.z-dn.net/?f=tan%5Ctheta%3D%20%5Cfrac%7Bopposite%20side%7D%7Badjacentside%7D.%20tan%2050%5E%7B%5Ccirc%7D%3D%5Cfrac%7BQR%7D%7B8.1%7D%2C%20tan50%5E%7B%5Ccirc%7D%20%3D%201.1917%2CQR%20%3D%20%281.1917%29%288.1%29%20%3D%209.65277)
SO QR is 9.65277 meters, rounding this off to the nearest tenth, we get QR = 9.7 inches which is option j.
Answer:
45 segundos.
Step-by-step explanation:
Un tren pasa por delante de un puente en 15 segundos; si el puente tuviera el doble de longitud, le tomaría el doble de tiempo en cruzarlo, si tuviera el triple de longitud, le tomaría el triple de tiempo y así sucesivamente.
En este caso, la pregunta es ¿En cuánto tiempo cruzaría el puente si tuviera el triple de su longitud? Por lo tanto, si cruza el puente en 15 segundos, teniendo el triple de longitud le tomaría 3 (15) = 45 segundos en cruzarlo.