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3 years ago

Please please help me. i’ll literally send you money. i need to pass please help

Mathematics

1 answer:

mihalych1998 [28]3 years ago

6 0

<h3>Answer: Approximately 6.4 units</h3>

==================================================

Explanation:

The origin is the point (0,0)

Use the distance formula to find the distance from (0, 0) to (4, -5)

Let

$(x_1,y_1) = (0,0)\\\\(x_2,y_2) = (4,-5)\\\\$

be our two points. Plug those values into the distance formula below and use a calculator to compute

$d = \sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\\\\d = \sqrt{\left(0-4\right)^2+\left(0-(-5)\right)^2}\\\\d = \sqrt{\left(0-4\right)^2+\left(0+5\right)^2}\\\\d = \sqrt{\left(-4\right)^2+\left(5\right)^2}\\\\d = \sqrt{16+25}\\\\d = \sqrt{41} \ \text{ exact distance}\\\\d \approx 6.40312 \ \text{ approximate distance}\\\\d \approx 6.4\\\\$

The distance between the two points (0,0) and (4,-5) is approximately 6.4 units.

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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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2 years ago

F(x+h) = (x+h)² + 3(x+h)^3<br><br> How would I distribute this?

Leno4ka [110]

Answer:

f(x + h) = 3x³ + x² + 9h²x + 3h³ + h² + 9hx² + 2hx

General Formulas and Concepts:

Order of Operations: BPEMDAS
Distributive Property
Expand by FOIL (First Outside Inside Last)
Combining like terms

Step-by-step explanation:

<u>Step 1: Define function</u>

f(x) = x² + 3x³

f(x + h) is x = x + h

<u>Step 2: Simplify</u>

Substitute: f(x + h) = (x + h)² + 3(x + h)³
Expand by FOILing: f(x + h) = (x² + 2hx + h²) + 3(x + h)³
Rewrite: f(x + h) = (x² + 2hx + h²) + 3(x + h)²(x + h)
Expand by FOILing: f(x + h) = (x²+2hx+h²) + 3(x² + 2hx + h²)(x+h)
Distribute/Expand: f(x + h) = (x²+2hx+h²) + 3(x³+3hx²+3h²x+h³)
Distribute 3: f(x + h) = (x²+2hx+h²)+(3x³+9hx²+9h²x+3h³)
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2 years ago

From the table below, determine whether the data shows an exponential function. Explain why or why not.

Ierofanga [76]

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3                                               1

1        1 - 3 = - 2                        2             2 - 1 = 1

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So, as you see y in increasing in regular constant intervals, when x also increases in regular constant intervals. => Δy / Δx = constant.

That is the result of a linear function, not an exponential one.

So, the true statement is given by the option <span>(A) No; the domain values are at regular intervals and the range values have a common sum 1. </span>

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3 years ago

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kakasveta [241]

The answer is: 1.49
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2 years ago

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EastWind [94]

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