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3 years ago

Help Please Is the following relation a function?

Mathematics

1 answer:

satela [25.4K]3 years ago

3 0

By definition a function cannot have a repeated x value. It can repeat the y value but not x. if you notice the x value of 0 is shared with the y values of 1 and 3. therefore this cannot be a function.

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PLEASE HELP ASAP 30 POINTS!!! WILL GIVE BRAINLIEST!!

r-ruslan [8.4K]

1. The factoring step is

sin²θ - cos²θ sin²θ = sin²θ (1 - cos²θ)

Then the Pythagorean identity is invoked:

cos²θ + sin²θ = 1 → 1 - cos²θ = sin²θ

so that

sin²θ - cos²θ sin²θ = sin²θ sin²θ = sin⁴θ

(third option)

2. Recall that a ² - b ² = (a - b) (a + b). The numerator here is such a difference of squares:

csc²x - 1 = (cscx - 1) (cscx + 1)

Then

(csc²x - 1) / (1 + sinx) = ((cscx - 1) (cscx + 1)) / (1 + sinx)

Recall that cscx = 1/sinx, so rewrite this as

… = ((1/sinx - 1) (1/sinx + 1)) / (1 + sinx)

In the numerator, pull out a factor of 1/sinx from both terms:

… = (1/sinx (1 - sinx) × 1/sinx (1 + sinx)) / (1 + sinx)

… = ((1 - sinx) (1 + sinx)) / (sin²x (1 + sinx))

Cancel the common factor of 1 + sinx :

… = (1 - sinx) / sin²x

Expand the fraction and rewrite sin in terms of csc :

… = 1/sin²x - sinx/sin²x

… = 1/sin²x - 1/sinx

… = csc²x - cscx

Factor out cscx to get the second option,

… = cscx (cscx - 1)

5 0

3 years ago

☺️☺️does anyone know☺️

Artemon [7]

The standard form of a circle:
$(x-h)^2+(y-k)^2=r^2$
Where:
(h; k) - the coordinates of a centerr - the radius
We have: $(x+5)^2+(y-10)^2=121$
therefore
$(-5;\ 10)$ - the center
$r^2=121\to r=\sqrt{121}\to r=11$ - the radius
Answer: 11

4 0

4 years ago

Read 2 more answers

Could u please solve it for me

Paraphin [41]

Hello there!

If x = -3

Calculate x^2 - x

We need to replace x by 3

3^2 - 3

=3 * 3 - 3

= 9 - 3

= 6

I hope this helps!

3 0

3 years ago

Jesse enjoys watching ‘Rudolph the Red-Nosed Reindeer’ during the holidays. If Jesse watches the movie three times every year, h

pogonyaev

Answer:

72