Answer:
The answers to the first three problems are shown in the figure attached
Fourth problem answer: 3.5 cm
Step-by-step explanation:
In problem 1, move the given triangle ABC four units to the right and 2 units down as what the displacement vector "v" indicates.
You may do such by translating each vertex of the triangle ABC such number of units one at a time and then joining the vertices.
In problem 2 the requested translation vector "v" indicates 4 units to the right and 1 unit up. Do such translation for each vertex of the triangle as suggested before.
In problem 3 the requested translation "v" asks for 2 units to the left and 3 up.
Do the translation of each vertex following these instructions.
Problem 4: use a ruler and notice that the length of the vector xy given has exactly the same length as the distance between the vertices A in one triangle, and A' in the other. The same is true for the distance between vertex B and B' in the other triangle, and for the distance between C and C'.
Answer:
7/20 is the scale factor
Step-by-step explanation:
We can see that Hexagon JKLMNO is bigger than Hexagon PQRSTU. This means that the scale factor is less than one.
LK is similar to QR.
QR/LK
7/20 is the scale factor.
Answer:
3n/8 +1=25
3n/8= 25-1
3n/8=24
Multiply both sides by 8 to clear the fraction on the left
3n=24x8
n=24x8/3
n=64
Answer and Step-by-step explanation:
THE QUESTION IS NOT FROM A QUIZ, IT IS FROM A HOMEWORK ASSIGNMENT.
<u>The unit rate is D. $4.00 per pound.</u>
This is because we see on the graph that at 1 pound of fudge, the cost is 4 dollars.
When looking at the unit rate, you want to look at y value for when x is 1.
<em><u>#teamtrees #PAW (Plant And Water)</u></em>
Answer:
EB=20, BC=8, AC=16
Step-by-step explanation:
The symbols indicate that:
AB=BC and AE=ED
EB and CD are parallels
AB=BC=8
AC= AB+BC
AC= 8+8
AC=16
To find EB we can use the Cosine Law
For the upper triangle x=∡EAB:
EB^2 = AB^2 + AE^2 -2*AB*AE*Cosx
AB*AE*Cosx= -(EB^2-AB^2 - AE^2)/2 (Part I)
For de big triangle:
DC^2= AC^2+AD^2 -2AC*AD*Cosx
Also:
AC=2*AB
AD=2*AE
DC^2= (2*AB)^2 + (2*AE)^2 -2(2*AB)(2*AE)*Cosx
DC^2= 4*AB^2 +4*AE^2- 8*AB*AE*Cosx
AB*AE*Cosx =-(DC^2-4*AB^2 -4*AE^2)/8 (Part II)
Part I= Part II
-(EB^2-AB^2 - AE^2)/2= -(DC^2-4*AB^2 -4*AE^2)/8
Extracting EB:
EB^2=DC^2/4
EB=DC/2
EB=40/2
EB=20
