All categories

3 years ago

On a map, 1/3 inch represents 25 miles. How many inches on this map represent 325 miles?

Mathematics

2 answers:

yulyashka [42]3 years ago

7 0

Answer:

4. 3 inches = 4.33333333333

Step-by-step explanation:

1/3 inch = 13 x 1/3 = 4.33 represents 325

As 1/3 x 1 = 3.33 represents 25

vampirchik [111]3 years ago

3 0

Answer:

4 $\frac{1}{3}$ inches

Step-by-step explanation:

You might be interested in

The amount of time a certain brand of light bulb lasts is normally distribued with a mean of 1800 hours and a standard deviation

DedPeter [7]

Answer: 68% of light bulbs last between 1765 hours and 1835 hours.

Step-by-step explanation:

The empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean . The empirical rule is further illustrated below

68% of data falls within the first standard deviation from the mean.

95% fall within two standard deviations.

99.7% fall within three standard deviations.

From the information given, the mean is 1800 hours and the standard deviation is 35 hours.

1 standard deviation = 1 × 35 = 35

1800 - 35 = 1765 hours

1800 + 35 = 1835 hours

Therefore, 68% of light bulbs last between 1765 hours and 1835 hours.

6 0

3 years ago

5. Estimate the value of √150 O between 5 and 6 O between 6 and 7 O between 7 and 8 O between 8 and 9

Digiron [165]

Answer:

basicaly your asking to be fulled so {{64

Step-by-step explanation:

3 0

3 years ago

What is the distance between the points (3, 4) ( -2 , 4)

rusak2 [61]

The answer is 5 using the distance formula

7 0

3 years ago

What is the volume of this rectangular pyramid?<br> 3 cm<br> 3 cm<br> 4 cm<br> cubic centimeters

Vsevolod [243]

Answer:

36 cm hope this helps

Step-by-step explanation:

3×3=9×4=36

5 0

2 years ago

Help me with differentation and integration please!!

Marina86 [1]

Answer:

See below

Step-by-step explanation:

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

Recall

$\dfrac{d}{dx}\tan x=\sec^2$

Using the chain rule

$\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}$

such that $u = \tan x$

we can get a general formulation for

$y = \tan^n x$

Considering the power rule

$\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}$

we have

$\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x$

therefore,

$\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x$

Now, once

$\sec^2 x - 1= \tan^2x$

we have

$3\tan^2x \sec^2x = 3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x$

Hence, we showed

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

================================================

For the integration,

$\int \sec^4 x\, dx$

considering the previous part, we will use the identity

$\boxed{\sec^2 x - 1= \tan^2x}$

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx$

Considering $u = \tan x$

and then $du=\sec^2x\ dx$

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx = \dfrac{\tan^3 x}{3}+\tan x + C }$

6 0

2 years ago