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tensa zangetsu [6.8K]
1 year ago
14

An elephant can run one over four mile in 36 seconds. Which of the following correctly shows this rate as miles per hour?

Mathematics
2 answers:
Vesnalui [34]1 year ago
7 0
It would be thirty six seconds over one-fourth mile= 144 miles per hour
ankoles [38]1 year ago
6 0

Answer: 1/4 mile over 36 seconds = 25 miles

Step-by-step explanation:

60 minutes x 60 seconds = 3600 ÷ 36 = 100  x  1/4 = 100 x .25= 25 miles

(1/4 mile = .25 because, .25 mile x 4 = 1 mile

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The time, in minutes, it took each of 11 students to complete a puzzle was recorded and is shown in the following list
bija089 [108]

Answer:

30, 31, 32, 35, and 38

Step-by-step explanation:

4 0
2 years ago
What is the domain and range of the function y = xn if n is an odd whole number? if n is an even whole number?
sergij07 [2.7K]

Answer:

Step-by What is the domain and range of the function y = xn if n is an odd whole number? if n is an even whole number?-step explanation:

8 0
2 years ago
Lin knows that there are 4 quarts in a gallon. She wants to convert 6 quarts to a gallon, but cannot decide if she should multip
defon

Given:

There are 4 quarts in a gallon.

She wants to convert 6 quarts to a gallon.

To find:

Whether she should multiply 6 by 4 or divide 6 by 4 to find her answer.

Solution:

We have,

4\text{ quarts}=1\text{ gallon}

1\text{ quarts}=\dfrac{1}{4}\text{ gallon}

6\text{ quarts}=\dfrac{6}{4}\text{ gallon}

Here, 6 quarts is equal to \dfrac{6}{4} gallon. It means we need to multiply 6 by 4 in 4\text{ quarts}=1\text{ gallon} to find the answer.

5 0
1 year ago
Deal with these relations on the set of real numbers: R₁ = {(a, b) ∈ R² | a > b}, the "greater than" relation, R₂ = {(a, b) ∈
uranmaximum [27]

Answer:

a) R1ºR1 = R1

b) R1ºR2 = R1

c) R1ºR3 = \{ (a,b) \in R^2 \}

d) R1ºR4 = \{ (a,b) \in R^2 \}

e) R1ºR5 = R1

f) R1ºR6 = \{ (a,b) \in R^2 \}

g) R2ºR3 = \{ (a,b) \in R^2 \}

h) R3ºR3 = R3

Step-by-step explanation:

R1ºR1

(<em>a,c</em>) is in R1ºR1 if there exists <em>b</em> such that (<em>a,b</em>) is in R1 and (<em>b,c</em>) is in R1. This means that a > b, and b > c. That can only happen if a > c. Therefore R1ºR1 = R1

R1ºR2  

This case is similar to the previous one. (<em>a,c</em>) is in R1ºR2 if there exists <em>b</em> such that (<em>a,b</em>) is in R2 and (<em>b,c</em>) is in R1. This means that a ≥ b, and b > c. That can only happen if a > c. Hence R1ºR2 = R1

R1ºR3

(a,c) is in R1ºR3 if there exists b such that a < b and b > c. Independently of which values we use for a and c, there always exist a value of b big enough so that b is bigger than both a and c, fulfilling the conditions. We conclude that any pair of real numbers are related.

R1ºR4

This is similar to the previous one. Independently of the values (a,c) we choose, there is always going to be a value b big enough such that a ≤ b and b > c. As a result any pair of real numbers are related.

R1ºR5

If a and c are related, then there exists b such that (a,b) is in R5 and (b,c) is in R1. Because of how R5 is defined, b must be equal to a. Therefore, (a,c) is in R1. This proves that R1ºR5 = R1

R1ºR6

The relation R6 is less restrictive than the relation R3, if we find 2 numbers, one smaller than the other, in particular we find 2 different numbers. If we had 2 numbers a and c, we can find a number b big enough such that a<b and b >c. In particular, b is different from a, so (a,b) is in R6 and (b,c) is in R1, which implies that (a,c) is in R1ºR6. Since we took 2 arbitrary numbers, then any pair of real numbers are related.

R2ºR3

This is similar to the case R1ºR3, only with the difference that we can take b to be equal to a as long as it is bigger than c. We conclude that any pair of real numbers are related.

R3ºR3

If a and c are real numbers such that there exist b fulfilling the relations a < b and b < c, then necessarily a < c. If a < c, then we can use any number in between as our b. Therefore R3ºR3 = R3

I hope you find this answer useful!

5 0
2 years ago
Pq=6x+25 and qr=16-3x; find pr
Viktor [21]

Given:\\\overline{PQ}=6x+25\\\overline{QR}=16-3x\\\\Finding:\overline{PR}

If \; Q \; is \; a \; point \; in \; between \; the \; line \; segment \; \overline{PR}, \; \\ then \; the \; distance \; of \; line \; segment \; \overline{PR} \; \\ can \; be \; found \; by \; adding \; line \; segment \; \overline{PQ} \; and \; \overline{QR}.

Using \; the \; concept \; \overline{PQ}+\overline{QR}=\overline{PR} \;,\\ we \; need \; to \; set \; up \; an \; equation \; given \; below:

\overline{PR} =6x+25 +16-3x\\\\Step \; 1: Grouping \; Like \; Terms\\\overline{PR} =6x-3x+25 +16\\\\Step \; 2: Combining \; Like \; Terms\\\overline{PR} =3x+41

Conclusion:

\overline{PR} \; is \; represented \; the \; expression \; 3x +41

3 0
1 year ago
Read 2 more answers
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