Ask question

Ask question

All categories

Oksi-84 [34.3K]

3 years ago

10

How to solve similar integer

1 answer:

SCORPION-xisa [38]3 years ago

6 0

Ok, I'll make it simple for you: Keep Change Change. For all subtraction problems (and only subtraction, no addition), all you have to do is keep the first number, then change the sign from a minus to a plus, and then change the last number to the opposite. For example:
7 - (-3)
7 + 3=10

-4 - 10
-4 + -10= -14

For addition, if both numbers are negative, you just add them together. For example:
-21 + (-13)= -34

In this case you switch the sign and the number's sign also:
7 + (-18)
7 - 18 = -11

I hope I was help!!!

You might be interested in

Solve this equation for x by graphing.<br> -4x - 1 = 5^x + 4 <br> Please help! Asap

Lina20 [59]

Answer:

-1.28

Step-by-step explanation:

You would put these two equations into a graphing ccalculator:

y = -4x - 1

and

y= 5^x + 4

Wherver they intersect is your answer.

These two happen to intersect at (-1.28,4.13)

Since the equation has x-values, you would give the first number (x-coordinate) as your answer.

7 0

3 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

3 years ago

(5x10^10) (5x10^5) in scientific notation

ss7ja [257]

Answer:

2.5e+16

Step-by-step explanation:

8 0

3 years ago

If angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 3:7:6:4, then name the type of quadrilateral

lesya692 [45]

Answer:whats the full questions?

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

helppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp

Klio2033 [76]

Answer:

12a + 18 = 12a + 18

Step-by-step explanation:

times what's in the brackets by what's to the left of the brackets

3 0

3 years ago