Answer:
1/3 (one-third)
Step-by-step explanation:
A scale factor that produces an image <u>smaller</u> than the original figure is <u>less</u> than 1. A scale factor that produces an image <u>bigger</u> than the original figure is <u>more</u> than 1.
Thus, the only option that's less than 1 is one-third.
Answer: 40
=====================================
Explanation:
The angle we want is QPR (bottom left), which is one of the base angles. The other base angle is QRP (bottom right). These two angles are equal because PQR is an isosceles triangle (PQ = RQ)
So if we can find angle QRP, then we have found angle QPR
Note how angle QRP and the 140 degree angle combine to form a straight 180 degree angle. Therefore these two angles add to 180 degrees
(angle QRP) + (140) = 180
(angle QRP) + 140 - 140 = 180-140 ... subtract 140 from both sides
angle QRP = 40
Since,
angle QPR = angle QRP
this means
angle QPR = 40
and
b = 40
Answer:(0,6)
Step-by-step explanation:
Solve for Y if X =0
y=2x+6
Y=2(0)+6
Y=6
X = -3, 3, 2i, -2i
Hope this helps! :)
Answer:
Option B is correct.i.e.,
Step-by-step explanation:
Given: Pyramid with equilateral triangle as base
Length of side of equilateral triangle = s unit
to find: height of equilateral triangle
Here we use a property of equilateral triangle.
Perpendicular from a vertex on a side and median of that side of a triangle is same in equilateral triangle.
All heights are of equal length. So, we just need to find one height or length of 1 altitude.
Figure of base triangle is attached
In Δ ABC
AB = BC = AC = s unit
AD is height
BD = 
Now, In Δ ABD
using pythagoras theorem
BD² + AD² = AB²








Therefore, Option B is correct.i.e., 