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2 years ago

Plz help PRE CALC !!!!!!!

Mathematics

1 answer:

sergey [27]2 years ago

5 0

Answer:

I think it's the second one. 5.6487

Step-by-step explanation:

You might be interested in

Whats the mid point of (-6,4) and (4,8)

mars1129 [50]

Answer:(-1, 5)

Step-by-step explanation:

Add X1 + X2 then divide by 2. Add Y1 + Y2 then divide by 2.

6 0

3 years ago

PLSSSS HELLPPPP MEEEEE I WILLL GIVEEE BRAINLIESSSTTT!!!!

earnstyle [38]

Answer:

$\displaystyle m=3$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Slope Formula: $\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Point (6, 8)

Point (5, 5)

Step 2: Find slope m

Simply plug in the 2 coordinates into the slope formula to find slope m

Substitute in points [SF]: $\displaystyle m=\frac{5-8}{5-6}$
[Fraction] Subtract: $\displaystyle m=\frac{-3}{-1}$
[Fraction] Divide: $\displaystyle m=3$

6 0

2 years ago

Read 2 more answers

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

Determine if the graph is symmetric about the x-axis, the y-axis, or the origin. r = 2 cos 3θ

Nina [5.8K]

Take a quick peek at brainly.com/question/10417591
that's how you test for the y-axis, x-axis and origin

recall your symmetry identities, cos(- θ) = cos(θ)

so, simply replace θ for those values, and if the resultant function looks exactly like the original, then it has that symmetry.

5 0

2 years ago

In a classroom, for every 8 girls there are 5 boys. What is the ratio of boys to girls?

MrRa [10]

Answer:

5:8

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers