A dilation is a transformation
, with center O and a scale factor of k that is not zero, that maps O to itself and any other point P to P'.
The center O is a fixed point, P' is the image of P, points O, P and P' are on the same line.
In a dilation of
, the scale factor,
is mapping the original figure to the image in such a way that the
distances from O to the vertices of the image are half the distances
from O to the original figure. Also the size of the image is half the
size of the original figure.
Therefore, <span>If
is a dilation of △ABC, the truth about the image △A'B'C'</span> are:
<span>AB is parallel to A'B'.
The distance from A' to the origin is half the distance from A to the origin.</span>
Answer:
a = 5 and b = 12
Step-by-step explanation:
<u>Step 1: Find angle B</u>
<em>Angle C = 90°</em>
<em>Angle A = 22.6°</em>
<em>Angle B = B</em>
<em>All angles in a triangle are equal to 180°.</em>
Angle A + Angle B + Angle C = 180°
22.6 + 90 + B = 180°
B = 180 - 112.6
B = 67.4°
<u>Step 2: Find the value of side AC 'b'</u>
<em>Hypotenuse = 13</em>
<em>Adjacent = b</em>
<em>Angle A = 22.6°</em>
Cos (Angle) = Adjacent/Hypotenuse
Cos (22.6) = b/13
b = 12
<u>Step 3: Find the value of side CB 'a'</u>
<em>Hypotenuse = 13</em>
<em>Opposite = a</em>
<em>Angle A = 22.6°</em>
Sin (angle) = Opposite/Hypotenuse
Sin (22.6°) = a/13
a = 4.99 rounded off to 5
Therefore, the value of a=5 and b=12.
!!
Answer:
∴Third side = 12 units
Step-by-step explanation:
In ΔCAB, the measure of segment AB is 8 units and the measure of segment AC is 5 units.
The sum of two sides of a triangle always grater than third side.
Therefore,
(5+8)=13 units
third side<13
∴Third side = 12 units
Answer:
4 in 2,417
4 in 2,147
Step-by-step explanation:
the place value of 4 in 2,417 is HUNDREDS
while the place value of 4 in 2,147 is TENS
Therefore the place value of 4 in 2,417 is ten times the value of 4 in 2,147
Let us make a list of all the details we have
We are given
The cost of each solid chocolate truffle = s
The cost of each cream centre chocolate truffle = c
The cos to each chocolate truffle with nuts = n
The first type of sweet box that contains 5 each of the three types of chocolate truffle costs $41.25
That is 5s+5c+5n = 41.25 (cost of each type of truffle multiplied by their respective costs and all added together)
The second type of sweet box that contains 10 solid chocolate trufles, 5 cream centre truffles and 10 chocolate truffles with nuts cost $68.75
That is 10s+5c+10n = $68.75
The third type of sweet box that contains 24 truffles evenly divided that is 12 each of solid chocolate truffle and chocolate truffle with nuts cost $66.00
That is 12s+12n=$66.00
Hence option C is the right set of equations that will help us solve the values of each chocolate truffle.
