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3 years ago

Only need help on number 1

Mathematics

1 answer:

Paraphin [41]3 years ago

5 0

Answer:

It can't

Step-by-step explanation:

I cannot represent values of a funcion because if you convert all the sets into fractions, (if simplified) then they will not be the same fraction. This is needed if you want a function that works.

The fractions are: 1/7, 3/16, 5/19, 1/4, 9/28. if simplified, then these are different fractions.

Additionally, you can see that it has both 5/19 and 5/20. (1/4).

If this was a working function it would yield the same output every time you put in 5 as an input.

Therefore, it cannot represent a function.

HAVE A GREAT DAY :D

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The vertex of parabola is the point at the intersection of parabola and its line of symmetry.

Now the given function is,

f(x) = 2x^2 – x – 6

Also, it is given that the vertex is located at (0.25, -6) and the parabola opens up, the function has two x-intercepts.

Comparing the given function with standard form,

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By comprison we get,

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Now, x-coordinate of vertex is given as,

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put the values we get,

x = -(-1)/2*2

or, x = 1/4

Put the value of x in given function, so y-coordinate of the vertex is given as,

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The vertex of the function is (one-quarter, negative 6 and one-eighth).
The function has two x-intercepts.

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In triangle JKL, sin(b°) = three fifths and cos(b°) = four fifths. If triangle JKL is dilated by a scale factor of 2, what is ta

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Answer:

Tan(b°) = 3/4 = three fourths (C)

Step-by-step explanation:

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Cos(b°) = 4/5

Tan(b°) = 3/4

From the above we know the value of each sides of ∆JKL

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Meaning we would multiply each sides of ∆JKL by 2. The values of the new triangle become:

Opposite = 2(3) = 6

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To find tan(b°) of the new triangle, we would apply tangent ratio

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Find attached the diagram for the new triangle (2)

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Answer:

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