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3 years ago

Where's a numerator found?

Mathematics

2 answers:

Sonja [21]3 years ago

4 0

In a fraction the numerator is the top of the fraction.

$\sf\frac{numerator}{denominator}$

lions [1.4K]3 years ago

3 0

A numerator is the number above a line of a common fraction.

For example: In 2/3, the numerator is 2.

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The sum of the first and second of three consecutive even integers is 158. find the three even integers.

Evgesh-ka [11]

Teh three terms are 78, 80, and 82

7 0

4 years ago

Slope Intercept<br> Graph each equation.<br> 1. y = 2x +3

liq [111]

Answer:

HERE U GO

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Area of a triangle = 1/2 • b•h

makvit [3.9K]

Answer:

Step-by-step explanation:

5*6=30

1/2*30=15

4 0

3 years ago

Daca a + b = 20 si b + c = 30 ,calculati 3 . a + 7 . b + 4 . c ( Dau coroana la cine ma ajuta )

Oduvanchick [21]

It is given that

a+b=20 -------------------- (1)

b+c=30 -----------------------(2)

We have to calculate

3 a + 4 b +7 c

Now multiplying equation(1 )by 3,we get

3 a+ 3 b=60 ----------------------------------(3)

and Multiplying equation( 2) by 4,we get

4 b + 4 c=120 -----------------------------------(4)

Adding expression (3) and (4),we get i.e left hand side of 3 to left hand side of 4 and right hand side of 3 to right hand side of 4.

3 a+ 3 b+ 4 b+ 4 c=60+120

Adding like terms, we get

3 a+ 7 b+ 4 c =180, Which is the required solution.

4 0

3 years ago

The radius of a right circular cylinder is given by √(t+6) and its height is 1/6√t , where t is time in seconds and the dimensio

denis23 [38]

Answer:

The rate change of volume of the cylinder is $\frac{\pi}{4} ( \sqrt t+\frac2{ \sqrt t})$ cubic inch per second.

Step-by-step explanation:

Given that the radius of right circular cylinder is $\sqrt{(t+6)}$ and its height is $\frac16 \sqrt t$ where t is time in second and the dimension are inches.

$\therefore r = \sqrt{(t+6)}$

The base area of the cylinder is A= $\pi r^2$

$=\pi (\sqrt{t+6})^2$

$= \pi (t+6)$

$\therefore A= \pi(t+6)$

Differentiating with respect to t

$\frac{dA}{dt}=\pi$

$\therefore h=\frac16\sqrt t$

Differentiating with respect to t

$\frac{dh}{dt}=\frac16 \times \frac12(t)^{\frac12-1}$

$\Rightarrow \frac{dh}{dt}=\frac1{12} (t)^{-\frac12}$

The volume of cylinder is V= Ah

∴V= Ah

Differentiating with respect to t

$\frac{dV}{dt}=A\frac{dh}{dt}+h\frac{dA}{dt}$

$=\pi (t+6). \frac1{12}t^{-\frac12} +\frac16\sqrt t . \pi$

$=\pi. \frac1{12}.t^{\frac12}+\pi . 6.\frac1{12} t^{-\frac12} +\pi\frac16 \sqrt t$

$=\frac{\pi}{4} ( \sqrt t+\frac2{ \sqrt t})$

The rate change of volume of the cylinder is $\frac{\pi}{4} ( \sqrt t+\frac2{ \sqrt t})$ cubic inch per second.

5 0

3 years ago