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1 year ago

Find the output, yyy, when the input, xxx, is 555. y = 5x-3y=5x−3

Mathematics

2 answers:

makkiz [27]1 year ago

4 0

The output y is $y=22$

Explanation:

The expression is $y=5 x-3$

We need to determine the output y, when the value of input is x is 5.

That is, substituting $x=5$ in $y=5 x-3$ , we get,

$y=5(5)-3$

Multiplying the value within the bracket, we have,

$y=25-3$

Subtracting the terms, we get,

$y=22$

Thus, the value of output y is $y=22$

Bond [772]1 year ago

3 0

Answer:

Step-by-step explanation:

its 7 on khan

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2 years ago

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A researcher in environmental science is conducting a study to investigate the impact of a particular herbicide on fish. He has

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Answer:

The herbicide has no impact on fish

Step-by-step explanation:

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1 year ago

Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:

professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is $\{1,2,3,4\}$ .

a) We need to find a relation R reflexive and transitive that contain the relation $R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}$

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

1R4 and 4R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
4R1 and 1R2, then 4 must be related with 2.

Therefore $\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\}$ is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

since 1R2, then 2 must be related with 1.
since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

1R2 and 2R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
2R1 and 1R4, then 2 must be related with 4.
4R1 and 1R2, then 4 must be related with 2.
2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

1 must be related with 1,

2 must be related with 2,

3 must be related with 3,

4 must be related with 4

For be symmetric

since 1R2, 2 must be related with 1,
since 1R4, 4 must be related with 1.

For be transitive

Since 4R1 and 1R2, 4 must be related with 2,
since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

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1 year ago

Joey wants to build a rectangular garden. He plans to use a side of a river for one side of the garden, he will not place fencin

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We let x and y be the measures of the sides of the rectangular garden. The perimeter subtracted with the other side should be equal to 92.

2x + y = 92

The value of y in terms of x is equal to,

y = 92 – 2x

The area is the product of the two sides,

A = xy

Substituting,

A = x (92 – 2x) = 92x – 2x2

Solving for the derivative and equating to zero,

0 = 92 – 4x ; x = 23

Therefore, the area of the garden is,

A = 23(92 – 2(23)) = 1058 yard2

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2 years ago

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