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2 years ago

Pls help!

Mathematics

1 answer:

Misha Larkins [42]2 years ago

5 0

Answer:

5.184 or 5.180 (rounded up)

Step-by-step explanation:

V = Bh or l x w x h

2.4 x 1.2 x 1.8 = 5.184 but rounded up would be 5.180

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How do you solve this problem?

umka21 [38]

The answer to your math question is x=5

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1 year ago

Graph the linear equation find three points that solve the equation then plot on the graph -3y=-x-7

Ira Lisetskai [31]

$- 3y = - x - 7 \\ y = \frac{1}{3} x + \frac{7}{3}$
$\binom{ - 1}{2} \: \: \: \: \: \: \binom{2}{3} \: \: \: \: \: \: \binom{5}{4}$
this is the line

7 0

3 years ago

Read 2 more answers

It takes 4 hours and 15 minutes to fly from Orlando, Florida, to Boston, Massachusetts. The distance between the two cities is 1

3241004551 [841]

Answer:

262.12mph and 419kph

Step-by-step explanation:

Given: It takes 4 hours and 15 minutes to fly from Orlando, Florida, to Boston, Massachusetts. The distance between the two cities is 1114 miles

To Find: the average speend of the plane in miles per hour, If every mile is approximately 1.6 kilometers, the speed of the airplane in kilometers per hour

Solution:

Distance between Orlando,Florida and Boston, Massachusetts $=1114$ $\text{miles}$

Time taken to cover the distance $=4\text{hours}$ and $15\text{minutes}$

We know that,

$\text{Average Speed}=\frac{\text{Total Distance}}{\text{Total time}}$

$=\frac{1114}{\frac{17}{4}}$

$=\frac{1114\times4}{17}$

$=262.12$ $\text{mph}$

It is given that,

$1\text{mile}=1.6\text{kilometer}$

therefore,

$1\text{mph}=1.6\text{kph}$

Speed Of plane in Kilometers per hours $=\text{speed in mph}\times1.6$

$419.4\text{kph}$

Speed of in miles per hour is $262.12$ $\text{mph}$ and Speed in kilometer per hour $419.4\text{kph}$

3 0

3 years ago

II The small capillaries in the lungs are in close contact with the alveoli. A red blood cell takes up oxygen during the 0.5s th

dangina [55]

The diffusion time of the red blood cell to take oxygen through the membrane is 25uS.

The small capillaries in the lungs are in close contact and it takes a red blood cell 0.5 seconds to squeeze through the capillary.

The diffusion time for the oxygen across the 1 um thick membrane separating the air is given by,

t = x²/2D

Where,

t is the diffusion time,

x is the thickness of the membrane,

D is the diffusion coefficient.

Putting values,

t = (1 x 10⁻⁹)²/2 x 2 x 10⁻¹¹

t = 10⁻⁷/4

t = 25 x 10⁻⁹ seconds.

t = 25 uS.

so, the diffusion time is 25uS.

To know more about Diffusion, visit,

brainly.com/question/94094

#SPJ4

3 0

1 year ago

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0

2 years ago