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3 years ago

Please assist me with the domain and range of graphs

Mathematics

1 answer:

Sphinxa [80]3 years ago

4 0

Answer: see below

Step-by-step explanation:

Domain represents the x-values from the smallest (furthest left) to the biggest (furthest right).

Range represents the y-values from the lowest (furthest down) to the highest (furthest up).

Interval notation: If a value is included (closed dot), use a bracket [ ]

If a value is NOT included (open dot), use a parenthesis ( )

Note that ± ∞ is never included.

1. Domain: smallest x-value is -4 (included). biggest x-value is 3 (included)

Range: lowest y-value is 2 (included). highest y-value is 5 (included)

D: x = [-4, 3] R: y = [2, 5]

2. Domain: smallest x-value is -∞ (←). biggest x-value is 2 (included)

Range: lowest y-value is 2 (included). highest y-value is ∞ (↑)

D: x = (-∞, 2] R: y = [2, ∞)

3. Domain: smallest x-value is -∞ (←). biggest x-value is ∞ (→)

Range: lowest y-value is -∞ (↓). highest y-value is ∞ (↑)

D: x = (-∞, ∞) R: y = (-∞, ∞)

4. Domain: smallest x-value is -∞ (←). biggest x-value is ∞ (→)

Range: lowest y-value is -∞ (↓). highest y-value is 4 (included)

D: x = (-∞, ∞) R: y = (-∞, 4]

5. Domain: smallest x-value is -∞ (←). biggest x-value is ∞ (→)

Range: lowest y-value is 3 (included). highest y-value is ∞ (↑)

D: x = (-∞, ∞) R: y = [3, ∞)

6. Domain: smallest x-value is -2 (included). biggest x-value is ∞ (→)

Range: lowest y-value is -3 (included). highest y-value is ∞ (↑)

D: x = [-2, ∞) R: y = [-3, ∞)

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Answer:

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• A = arcsin(8/√89) ≈ 57.995°

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Step-by-step explanation:

By the law of cosines, ...

c² = a² + b² -2ab·cos(C)

Since c=90°, cos(C) = 0 and this reduces to the Pythagorean theorem for this right triangle.

c = √(8² +5²) = √89 ≈ 9.434

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sin(A) = a/c·sin(C) = a/c = 8/√89

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Answer:

Thus, the two root of the given quadratic equation $x^2+4=6x$ is 5.24 and 0.76 .

Step-by-step explanation:

Consider, the given Quadratic equation, $x^2+4=6x$

This can be written as , $x^2-6x+4=0$

We have to solve using quadratic formula,

For a given quadratic equation $ax^2+bx+c=0$ we can find roots using,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ...........(1)

Where, $\sqrt{b^2-4ac}$ is the discriminant.

Here, a = 1 , b = -6 , c = 4

Substitute in (1) , we get,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\Rightarrow x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot 1 \cdot (4)}}{2 \cdot 1}$

$\Rightarrow x=\frac{6\pm\sqrt{20}}{2}$

$\Rightarrow x=\frac{6\pm 2\sqrt{5}}{2}$

$\Rightarrow x={3\pm \sqrt{5}}$

$\Rightarrow x_1={3+\sqrt{5}}$ and $\Rightarrow x_2={3-\sqrt{5}}$

We know $\sqrt{5}=2.23607$ (approx)

Substitute, we get,

$\Rightarrow x_1={3+2.23607}$ (approx) and $\Rightarrow x_2={3-2.23607}$ (approx)

$\Rightarrow x_1={5.23607}$ (approx) and $\Rightarrow x_2=0.76393}$ (approx)

Thus, the two root of the given quadratic equation $x^2+4=6x$ is 5.24 and 0.76 .

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PROOF:

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3 years ago

Please assist me with the domain and range of graphs​

Please assist me with the domain and range of graphs