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Paraphin [41]
2 years ago
13

If f (x) = startroot one-half x minus 10 endroot 3, which inequality can be used to find the domain of f(x)? startroot one-half

x endroot greater-than-or-equal-to 0 one-half x greater-than-or-equal-to 0 one-half x minus 10 greater-than-or-equal-to 0 startroot one-half x minus 10 endroot 3 greater-than-or-equal-to 0
Mathematics
1 answer:
tankabanditka [31]2 years ago
6 0

A function assigns values. The inequality one-half x minus 10 greater-than-or-equal-to 0 should be satisfied.

<h3>What is a Function?</h3>

A function assigns the value of each element of one set to the other specific element of another set.

The domain of a function is a set of values for which the equation is satisfied. Since the function has under root values, the value which is under the root should be non-negative so that it does not give a complex number,  therefore,

The inequality \sqrt{\dfrac12x^2-10}\geq 0satisfied to find the domain of the function f(x).

Also, we can write that 0.5x²-10≥0 for the value to be positive.

Hence, the inequality one-half x minus 10 greater-than-or-equal-to 0 should be satisfied.

Learn more about Function:

brainly.com/question/5245372

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Step-by-step explanation:

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Can someone please help me with these problems and show work, if the answer isn’t with work I will remove your comment and you w
MrRissso [65]

m∠B = 57.52°, m∠B = 70.8°, AB = 46.03 km and AC = 39.08 ft. This can be obtained using the Laws of cosine formula and Laws of sine formula.

<h3>Find the required angles and sides:</h3>
  • Laws of cosine formula,

In a triangle ABC,

⇒ a² = b² + c² - 2bc cos A

⇒ b² = a² + c² - 2ac cos B

⇒ c² = a² + b² - 2ab cos C

where a, b and c are sides of a triangle and A, B and C are the angles of a triangle.

  • Laws of sine formula,

In a triangle ABC,

⇒ \frac{sinA}{a} =\frac{sinB}{b} =\frac{sinC}{c}

where a, b and c are sides of a triangle and A, B and C are the angles of a triangle.

 

In the question we can use the laws of cosine formula and laws of sine formula,

5) Given that,

AB = c = 13 km

AC = b = 21 km

m∠A = 91°

By using Laws of cosine formula,

⇒ a² = b² + c² - 2bc cos A

a² = 21² + 13² - 2(21)(13) cos 91°

a² = 441 + 169 - 546 (-0.0174524064)    

a² = 610 + 9.52901391 = 619.529014

⇒ a = 24.89 km

By using Laws of sine formula,

⇒ \frac{sinA}{a} =\frac{sinB}{b} =\frac{sinC}{c}

sin 91°/24.89 = sin B/21

sin B = 0.999847695×21/24.89 = 0.843583833

⇒ m∠B = 57.52°

 

6) Given that,

AB = c = 11 cm

BC = a = 13 cm

AC = b = 14 cm

By using Laws of cosine formula,

⇒ b² = a² + c² - 2ac cos B

14² = 13² + 11² - 2(13)(11) cos B

196 = 169 + 121 - 286 cos B

196 = 290 - 286 cos B

cos B = 94/286

cos B = 0.328671329

⇒ m∠B = 70.8°

 

7) Given that,

AC = b = 24 km

BC = a = 26 km

m∠C = 134°

By using Laws of cosine formula,

⇒ c² = a² + b² - 2ab cos C

c² = 26² + 24² - 2(26)(24) cos 134°

c² = 676 + 576 - 1248 (-0.69465837)

c² = 1252 + 866.933646 = 2118.93365

⇒ AB = c = 46.03 km

8) Given that,

AB = c = 26 ft

BC = a = 21 ft

m∠B = 112°

By using Laws of cosine formula,

⇒ b² = a² + c² - 2ac cos B

b² = 21² + 26² - 2(21)(26) cos 112°

b² = 441 + 676 - 1096 (-0.374606593)

b² = 1117 + 410.568826 = 1527.56883

⇒ AC = b = 39.08 ft

Hence m∠B = 57.52°, m∠B = 70.8°, AB = 46.03 km and AC = 39.08 ft.

Learn more about Laws of cosine and Laws of sine here:

brainly.com/question/17289163

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