Ask question

Ask question

All categories

3 years ago

10

Which of the following is an irrational number A.3\14 B.3.8 C.√38 D.√64

1 answer:

Vesna [10]3 years ago

6 0

It is
$\sqrt{38}$
because an irrational number cannot be written as a simple fraction ( because there is not a finite number of numbers when written as a decimal ).

You might be interested in

MARK BRAINLIESTTTTTT

Gre4nikov [31]

Answer:

i think its c hope it helps

Step-by-step explanation:

8 0

3 years ago

HELP I need this for a quiz the screenshot is there

Setler79 [48]

Answer:

The value is -30

Step-by-step explanation:

Plug -6 in for x in the equation:

3sqrt-6(-6)+4(-6)

-6-4*6

-6-24

-30

4 0

3 years ago

What is the equation of the line that passes<br> through the point (3, 4) and has a slope of -2/3

Rina8888 [55]

Answer:

y=-2/3x+6

Step-by-step explanation:

y-y1=m(x-x1)

y-4=-2/3(x-3)

y=-2/3x+6/3+4

y=-2/3x+2+4

y=-2/3x+6

6 0

2 years ago

A boy owns 7 pairs of pants, 3 shirts, 2 ties, and 6 jackets. How many different outfits can he wear to school if he must wear o

lutik1710 [3]

Answer:

252.

Step-by-step explanation:

We use the Multiplication Rule:

Number of outfits = 7 * 3 * 2 * 6 = 252.

<u />

3 0

2 years ago

Read 2 more answers

The radius of a cone is increasing at a constant rate of 7 meters per minute, and the volume is decreasing at a rate of 236 cubi

storchak [24]

Answer:

The rate of change of the height is 0.021 meters per minute

Step-by-step explanation:

From the formula

$V = \frac{1}{3}\pi r^{2}h$

Differentiate the equation with respect to time t, such that

$\frac{d}{dt} (V) = \frac{d}{dt} (\frac{1}{3}\pi r^{2}h)$

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (r^{2}h)$

To differentiate the product,

Let r² = u, so that

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (uh)$

Then, using product rule

$\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h\frac{du}{dt}]$

Since $u = r^{2}$

Then, $\frac{du}{dr} = 2r$

Using the Chain's rule

$\frac{du}{dt} = \frac{du}{dr} \times \frac{dr}{dt}$

∴ $\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h(\frac{du}{dr} \times \frac{dr}{dt})]$

Then,

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

Now,

From the question

$\frac{dr}{dt} = 7 m/min$

$\frac{dV}{dt} = 236 m^{3}/min$

At the instant when $r = 99 m$

and $V = 180 m^{3}$

We will determine the value of h, using

$V = \frac{1}{3}\pi r^{2}h$

$180 = \frac{1}{3}\pi (99)^{2}h$

$180 \times 3 = 9801\pi h$

$h =\frac{540}{9801\pi }$

$h =\frac{20}{363\pi }$

Now, Putting the parameters into the equation

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

$236 = \frac{1}{3}\pi [(99)^{2} \frac{dh}{dt} + (\frac{20}{363\pi }) (2(99)) (7)]$

$236 \times 3 = \pi [9801 \frac{dh}{dt} + (\frac{20}{363\pi }) 1386]$

$708 = 9801\pi \frac{dh}{dt} + \frac{27720}{363}$

$708 = 30790.75 \frac{dh}{dt} + 76.36$

$708 - 76.36 = 30790.75\frac{dh}{dt}$

$631.64 = 30790.75\frac{dh}{dt}$

$\frac{dh}{dt}= \frac{631.64}{30790.75}$

$\frac{dh}{dt} = 0.021 m/min$

Hence, the rate of change of the height is 0.021 meters per minute.

3 0

3 years ago