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3 years ago

Nine less than twice the height H

Mathematics

1 answer:

PIT_PIT [208]3 years ago

4 0

Answer:

2h-9

Step-by-step explanation:

Twice the height of h would be 2H and 9 less from that would be 2h-9

Hope this helps :) Plz mark brainliest <3

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Celine is estimating the number of tiles in a mosaic. There are 7 different colors, and 472 tiles of each color. There are also

ValentinkaMS [17]

Answer:

3,400 tiles

Step-by-step explanation:

Since there is 472 tiles of each 7 colors,

Multiply 472 by 7 to get 3,304

Then you would add the 96 blank tiles to get exactly

3.400 tiles.

8 0

3 years ago

The sum of 3 and 7 times a number is 45

san4es73 [151]

Answer:

4.5

Step-by-step explanation:

The sum of 3 and 7 is 10

then what number to you multiply 10 by to get 45

45 ÷ 10 = 4.5

so the answer is 4.5

6 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

David travels from Building A to Building B. He can either cycle

Lelechka [254]

Answer:

43 minutes

Step-by-step explanation:

<em>Let cycling speed be x and driving speed be y</em>

<u>Equations as per given, considering same distance in both cases:</u>

19x + 8y = 13x + 10y
19x - 13x = 10y - 8y
6x = 2y
3x = y

We see that driving speed is 3 times greater than cycling speed

Then 8 minutes driving = 8*3= 24 minutes of cycling, or 10 minutes of driving = 10*3= 30 minutes of cycling:

19 + 24 = 43 minutes or
13 + 10*3 = 43 minutes is the time to cycle between A and B

3 0

3 years ago

Two positive integers have a sum of 23 and a product of 126. Find the two numbers.

meriva

Answer:

SUBTRACT

Step-by-step explanation:

6 0

3 years ago