Answer:
✔️2 sets of corresponding angles
<D and <S
<R and <L
✔️2 sets of corresponding sides
DR and SL
RM and LT
Step-by-step explanation:
When two polygons are congruent, it implies that they have the same shape and size. Therefore, their corresponding angles and sides are congruent to each other.
When naming congruent polygons, the arrangement of the vertices are kept in a definite order of arrangement.
Therefore, Given that polygon DRMF is congruent to SLTO, the following angles and sides correspond to each other:
<D corresponds to <S
<R corresponds to <L
<M corresponds to <T
<F corresponds to <O
For the sides, we have:
DR corresponds to SL
RM corresponds to LT
MF corresponds to TO
FD corresponds to OS.
We can select any two out of these sets of corresponding angles and sides as our answer. Thus:
✔️2 sets of corresponding angles
<D and <S
<R and <L
✔️2 sets of corresponding sides
DR and SL
RM and LT
Answer:
z = 5*(1/2)
z = 5/10
---
time switching classes:
w = 7/10
---
y - 6x - z - w = 0
6x = y - z - w
x = (y - z - w)/6
x = (76/10 - 5/10 - 7/10)/6
x = (76 - 5 - 7)/(10*6)
x = (64)/(10*6)
x = (2*2*2*2*2*2)/(2*5*2*3)
x = (2*2*2*2)/(5*3)
x = 16/15
x = 1.0666666666
---
check:
y = 7 + 3/5
y = 7.6
z = 1/2
z = 0.5
w = 7/10
w = 0.7
y - 6x - z - w = 0
6x = y - z - w
x = (y - z - w)/6
x = (7.6 - 0.5 - 0.7)/6
x = 1.0666666666
---
answer:
z = 5*(1/2)
z = 5/10
---
time switching classes:
w = 7/10
---
y - 6x - z - w = 0
6x = y - z - w
x = (y - z - w)/6
x = (76/10 - 5/10 - 7/10)/6
x = (76 - 5 - 7)/(10*6)
x = (64)/(10*6)
x = (2*2*2*2*2*2)/(2*5*2*3)
x = (2*2*2*2)/(5*3)
x = 16/15
x = 1.0666666666
---
check:
y = 7 + 3/5
y = 7.6
z = 1/2
z = 0.5
w = 7/10
w = 0.7
y - 6x - z - w = 0
6x = y - z - w
x = (y - z - w)/6
x = (7.6 - 0.5 - 0.7)/6
x = 1.0666666666
---
answer:
each class is 1.07 hours
Step-by-step explanation:
Answer:
In algebraic terms this means x-17>33
Step-by-step explanation:
to solve it x−17>33
1 Add 17 to both sides.
x>33+17
2 Simplify 33+17 to 50
x>50
Answer:
=
+
+
Step-by-step explanation:
=multiple ways to climb a tower
When n = 1,
tower= 1 cm
= 1
When n = 2,
tower =2 cm
= 2
When n = 3,
tower = 3 cm
it can be build if we use three 1 cm blocks
= 3
When n = 4,
tower= 4 cm
it can be build if we use four 1 cm blocks
= 6
When n > 5
tower height > 4 cm
so we can use 1 cm, 2 cm and 4 cm blocks
so in that case if our last move is 1 cm block then will be
n —1 cm
if our last move is 2 cm block then will be
n —2 cm
if our last move is 4 cm block then will be
n —4 cm
=+ +
