Answer:
I can't include a picture but it will be the same shape except it will be flipped over and will look a bit like differently angled packman with the back of the mouth apart the same distance, with the mouth "pointier" part of it one line away from the red line from the other side.... Oof sorry if this didn't help, I tried my best...
Step-by-step explanation:
From the "point" of the blue triangle, go diagonally SE 2 spaces. From there, draw a straight line to the right for 3 spaces. From there, draw upwards two spaces, and then connect those two lines into a triangle.
Answer:
answer
Step-by-step explanation:
you have to subtract 3b from 3a so the answer is 0.
Triangle A is the original triangle, and triangle A' is the image.
We need a pair of corresponding sides in the two triangles that are both labeled with lengths.
We see the side labeled 15.2 mm in triangle A', and we see the side labeled 3.8 mm in triangle A. These two sides are corresponding sides.
Divide the length of the side in the image by the length of the side in the original to find the scale factor.
(15.2 mm)/(3.8 mm) = 4
The scale factor is 4.
Answer:
Port r is 100° from Port p and 26km from Port p
Step-by-step explanation:
Lets note the dimension.
From p to q = 15 km = a
From q to r = 20 km= b
Angle at q = 50° + 45°
Angle at q = 95°
Ley the unknown distance be x
Distance from p to r is the unknown.
The formula to be applied is
X²= a²+ b² - 2abcosx
X²= 15² + 20² - 2(15)(20)cos95
X²= 225+400-(-52.29)
X²= 677.29
X= 26.02
X is approximately 26 km
To know it's direction from p
20/sin p = 26/sin 95
Sin p= 20/26 * sin 95
Sin p = 0.7663
P= 50°
So port r is (50+50)° from Port p
And 26 km far from p
Answer:
3^2/5
Step-by-step explanation:
