A fraction reduces to 36. If its denominator is 6x^5, what it's its numerator?

2 answers:

6 0

The answer is B because you will cancel the x^5 part on top and bottom and the numerator left 6^3 and denominator left 6
6^3=216divid by 6 you will get 36

Inessa [10]3 years ago

6 0

Answer:

Option (b) is correct.

Numerator is $N=6^3x^5$ .

Step-by-step explanation:

Given : A fraction reduces to 36. If its denominator is $6x^5$ .

We have to find the numerator.

Let numerator be N.

and denominator = $6x^5$ .

Fraction (F) is numerator (N) over denominator(D).

$F=\frac{N}{D}$

Given : fraction is 36

$36=\frac{N}{6x^5}$

$6^2=\frac{N}{6x^5}$

Cross multiply , we get,

$(6^2)(6x^5)=N$

$N=6^3x^5$

Thus, numerator is $N=6^3x^5$ .

Option (b) is correct.

You might be interested in

What is the approximate diameter of a sphere with a volume of 34 cm

Nutka1998 [239]

Answer:

<h3>The diameter is 4cm</h3>

Step-by-step explanation:

Volume of a sphere is

$V = \frac{4}{3} \pi {r}^{3}$

Where r is the radius

diameter = radius × 2

To find the diameter we must first find the radius

Volume = 34cm³

That's

$34 = \frac{4}{3} \pi {r}^{3} \\ \\ 34 \times 3 = 4\pi {r}^{3} \\ \\ 102 = 4\pi {r}^{3} \\ {r}^{3} = \frac{102}{4} \pi \\ \\ r = \sqrt[3]{ \frac{51}{2\pi}} \\ \\ r = 2.01 \\ \\ r = 2.0 cm$

Diameter = 2 × 2cm

Hope this helps you

6 0

3 years ago

What is the vertex of a parabola defined by the equation <br> x = 5y2?

QveST [7]

Find the critical points of f(y):Compute the critical points of -5 y^2
To find all critical points, first compute f'(y):( d)/( dy)(-5 y^2) = -10 y:f'(y) = -10 y
Solving -10 y = 0 yields y = 0:y = 0
f'(y) exists everywhere:-10 y exists everywhere
The only critical point of -5 y^2 is at y = 0:y = 0
The domain of -5 y^2 is R:The endpoints of R are y = -∞ and ∞
Evaluate -5 y^2 at y = -∞, 0 and ∞:The open endpoints of the domain are marked in grayy | f(y)-∞ | -∞0 | 0∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:The open endpoints of the domain are marked in grayy | f(y) | extrema type-∞ | -∞ | global min0 | 0 | global max∞ | -∞ | global min
Remove the points y = -∞ and ∞ from the tableThese cannot be global extrema, as the value of f(y) here is never achieved:y | f(y) | extrema type0 | 0 | global max
f(y) = -5 y^2 has one global maximum:Answer: f(y) has a global maximum at y = 0