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Blababa [14]
3 years ago
13

The dimensions of Kara’s bedroom floor are 8 3/4 feet by 10 2/3. What if the area of her bedroom floor

Mathematics
1 answer:
olasank [31]3 years ago
6 0

Answer:

93\frac{1}{3}

Step-by-step explanation:

The area of a bedroom floor is the width multiplied by the length. We know that the length's is 10\frac{2}{3} feet and the width is 8\frac{3}{4} feet. We just multiply the 2 fractions to find the area of her bedroom floor in feet squared

10\frac{2}{3} * 8\frac{3}{4}

→ There area various methods but I am going to use the improper fraction method. First convert the fractions into improper fractions

\frac{32}{3} *\frac{35}{4}

→ We can cancel out 4 and 32 with 1 and 8

\frac{8}{3} *\frac{35}{1}

→ Multiply like normal fractions

\frac{280}{3}

→ Convert into mixed fraction

93\frac{1}{3}

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Juli2301 [7.4K]

Step-by-step explanation:

a+b+c=180°

64°+b+c=180°

b=c

2b+64°=180°

2b=180-64

2b=116°

b=116/2

b=58°=c

4 0
3 years ago
Tyler is thinking of a number that is divisible by 2 and by 3. Write another number in which Tyler's number must also be divisib
Mumz [18]

Answer:

6

Step-by-step explanation:

6 because 6/2=3,6/3=2

3 0
3 years ago
A circle has diameter 12. What's the circumference?
Karo-lina-s [1.5K]

Answer:

b. 12π

Step-by-step explanation:

C= 2πr

R= D/2= 12/2= 6

C= 2x3.14x6

2x6= 12

C= 12π

6 0
3 years ago
Read 2 more answers
Which of these relations on the set {0, 1, 2, 3} are equivalence relations? If not, please give reasons why. (In other words, if
Delicious77 [7]

Answer:

(1)Equivalence Relation

(2)Not Transitive, (0,3) is missing

(3)Equivalence Relation

(4)Not symmetric and Not Transitive, (2,1) is not in the set

Step-by-step explanation:

A set is said to be an equivalence relation if it satisfies the following conditions:

  • Reflexivity: If \forall x \in A, x \rightarrow x
  • Symmetry: \forall x,y \in A, $if x \rightarrow y,$ then y \rightarrow x
  • Transitivity: \forall x,y,z \in A, $if x \rightarrow y,$ and y \rightarrow z, $ then x \rightarrow z

(1) {(0,0), (1,1), (2,2), (3,3)}

(3) {(0,0), (1,1), (1,2), (2,1), (2,2), (3,3)}

The relations in 1 and 3 are Reflexive, Symmetric and Transitive. Therefore (1) and (3) are equivalence relation.

(2) {(0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3)}

In (2), (0,2) and (2,3) are in the set but (0,3) is not in the set.

Therefore, It is not transitive.

As a result, the set (2) is not an equivalence relation.

(4) {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3)}

(1,2) is in the set but (2,1) is not in the set, therefore it is not symmetric

Also, (2,0) and (0,1) is in the set, but (2,1) is not, rendering the condition for transitivity invalid.

5 0
3 years ago
Hope someone can help me with this
postnew [5]

The perimeter:

40+35+16+16+125.7= 232.7 m

btw lowkey frustrating that they alternate between cm and m


8 0
3 years ago
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