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Lana71 [14]
3 years ago
15

Did the vertical or horizontal asymptote change? If it didn’t change, why not?

Mathematics
1 answer:
PIT_PIT [208]3 years ago
7 0

Answer:

<em>I think it didnt change</em>

Step-by-step explanation:

<em>Cause both areas from the line are the same and if it were to change the left would be diffrent from the right. So there i think it didnt change.</em>

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On Monday, Torrance practiced the violin for 4/10 of an hour. On Tuesday, he practiced for 1 6/8 of an hour. How much time did h
Yuri [45]

Answer: 2\frac{3}{20}

<u>Step-by-step explanation:</u>

First, convert mixed numbers into improper fractions and simplify all fractions

\frac{4}{10} ÷ \frac{2}{2} = \frac{2}{5}

1\frac{6}{8} = \frac{8(1) + 6}{8} = \frac{14}{8} ÷ \frac{2}{2} = \frac{7}{4}

Next, find their sum.  Remember to find the LCD and convert the fractions so they have like denominators.

Monday + Tuesday = Total

\frac{2}{5} + \frac{7}{4} = Total      <em>the LCD of 5 and 4 is 20</em>

(\frac{4}{4})\frac{2}{5} + (\frac{5}{5})\frac{7}{4} = Total

\frac{8}{20} + \frac{35}{20} = Total

\frac{43}{20} = Total

Then, convert the improper fraction into a mixed number

\frac{43}{20} = 2\frac{3}{20}

7 0
3 years ago
There are 12 inches (in). in 1 foot (ft). which expression can be used to find the number of inches in 2.5 feet?
Ludmilka [50]
Given:
12 inches in 1 foot
number of inches in 2.5 feet

2.5 feet * 12 inches/1foot = 2.5 * 12 inches = 30 inches

1 foot + 1 foot + 0.5 foot = 2.5 feet
12 inches + 12 inches + 6 inches = 30 inches

12 inches : 1 foot = x : 2.5 feet
12 inches * 2.5 feet = x*1foot
30 inches * feet = x * foot
30 inches = x
3 0
3 years ago
The graphically and algebraically
sertanlavr [38]
Graphically is intersect i believe
6 0
2 years ago
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
Simplify -3i sqrt -40<br><br> Enter your answer in simplest radical form
Mariulka [41]

Answer:

Step-by-step explanation:

-3i  √(-40) = -3i √40i²    ...(i² = - 1)

                 = -3i √40 i

              -3i  √(-40)  =   3√40

6 0
3 years ago
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