Your answer would be option B. 2y² - y - 6 = 0. This is because if you were to substitute x = y² - 1 into the equation 2x - y = 4, you would get 2(y² - 1) - y = 4, which expands into 2y² - 2 - y = 4, and then simplifies to 2y² - y - 6 = 0.
I hope this helps!
Volume of a sphere and a cone
We have that the equation of the volume of a sphere is given by:

We have that the radius of a sphere is half the diameter of it:
Then, the radius of this sphere is
r = 6cm/2 = 3cm
<h2>Finding the volume of a sphere</h2>
We replace r by 3 in the equation:

Since 3³ = 3 · 3 · 3 = 27

If we use π = 3.14:

Rounding the first factor to the nearest hundredth (two digits after the decimal), we have:
4.18666... ≅ 4.19
Then, we have that:

Then, we have that:
<h2>Finding the volume of a cone</h2>
We have that the volume of a cone is given by:

where r is the radius of its base and h is the height:
Then, in this case
r = 3
h = 6
and
π = 3.14
Replacing in the equation for the volume:

Then, we have:
3² = 9

Answer: the volume of the cone that has the same circular base and height is 56.52 cm³
Answer:
Lets solve = 3/5 ÷ 1/2
3/5 ÷ 1/2
= 3/5 × 2/1
= 3 × 2/5 × 1
= 6/5
In mixed fraction its

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Now lets solve "a" option
3/5 × 2
= 3 × 2/5 × 1
= 6/5
In mixed fraction

So option a = 3/5 x 2/1 is ur answer
The numeric values for the given functions are as follows:
<h3>How to find the numeric value of a function or of an expression at a given point?</h3>
To find the numeric value of a function at x = a, we replace each instance of the variable, usually x, in the function by the desired value of a.
Function f(x) is defined by:
f(x) = x².
For the numeric value at x = 1/3, we replace the lone instance of x by 1/3, hence:
f(1/3) = (1/3)² = 1/9.
Function g(x) is defined by:
g(x) = 2x.
For the numeric value at x = 4, we replace the lone instance of x by 4, hence:
g(4) = 2(4) = 8.
For the numeric value at x = -3, we replace the lone instance of x by -3, hence:
g(-3) = 2(-3) = -6.
More can be learned about the numeric values of a function at brainly.com/question/28367050
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