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Brut [27]
3 years ago
15

If the function f(x)=3x^2-5 and has the domain {-4,-3,1,5}, what is its range?

Mathematics
1 answer:
Paraphin [41]3 years ago
8 0
The answer would be C. Hope the picture helps.
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part 2. Find the value of the trig function indicated, use for that Pythagorean theorem to find the third side if you need it.​
andreev551 [17]

Answer:  \bold{5)\ \cos \theta=\dfrac{\sqrt{11}}{6}\qquad 6)\ \tan \theta =\dfrac{8}{17}\qquad 7)\ \cos \theta = \dfrac{4}{3}\qquad 8)\ \cos \theta = \dfrac{\sqrt{10}}{10}}

<u>Step-by-Step Explanation:</u>

Pythagorean Theorem is: a² + b² = c²  , <em>where "c" is the hypotenuse</em>

5)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{3\sqrt{11}}{18}\quad \rightarrow \large\boxed{\dfrac{\sqrt{11}}{6}}

Note: (15)² + (3√11)² = hypotenuse²   →   hypotenuse = 18

6)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{8}{17}\quad =\large\boxed{\dfrac{8}{17}}

Note: 8² + 15² = hypotenuse²   →   hypotenuse = 17

7)\ \tan \theta=\dfrac{\text{side opposite to}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{20}{15}\quad \rightarrow \large\boxed{\dfrac{4}{3}}

Note: hypotenuse not needed for tan

8)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{2}{2\sqrt{10}}\quad =\large\boxed{\dfrac{\sqrt{10}}{10}}

Note: 2² + 6² = hypotenuse²   →   hypotenuse = 2√10

8 0
2 years ago
A pool table is 95 inches long and 51 inches wide. what is the length of the diagonal of the pool table?
anzhelika [568]
I really think your is 12 feet and 2 inches
3 0
3 years ago
469.4 \62 estimate the quotient
bulgar [2K]

469.4/62=7.57

that is the answer

7 0
3 years ago
2. Triangle ABC is equilateral and AB = 6x + 3, BC = 8x - 1,2. AB =
kherson [118]

Answer:

AB = BC = AC = 15

Step-by-step explanation:

Since the triangle is equilateral then the 3 sides are congruent.

Equate any 2 sides and solve for x , that is

AC = AB , so

11x - 7 = 6x + 3 ( subtract 6x from both sides )

5x - 7 = 3 ( add 7 to both sides )

5x = 10 ( divide both sides by 5 )

x = 2

Then

AC = 11x - 7 = 11(2) - 7 = 22 - 7 = 15 , thus

AB = BC = AC = 15

8 0
2 years ago
Expand and simplify 5(2a – c) + 4(3a + 2c)
SIZIF [17.4K]
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7 0
3 years ago
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