All categories

3 years ago

Find the vertex of the function h(x) = -x^2 + 14

Mathematics

1 answer:

Mazyrski [523]3 years ago

5 0

The vertex form:

$f(x)=a(x-h)^2+k\\\\(h,\ k)-vertex$

$h(x)=-x^2+14=-(x-0)^2+14\to h=0,\ k=14$

<h3>Answer: (0, 14).</h3>

You might be interested in

Solve the equation. 3(2x+2)=9x+2-3x+4

inessss [21]

Answer: x = 4

Step-by-step explanation:

6 0

3 years ago

One glue stick has 1/5 left and the other 3/10 left, which is bigger

dedylja [7]

The glue stick with 3/10

6 0

3 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

25(3x-4y)^2-40(9x^2-16y^2)+16(3x+4y)^2

Rama09 [41]

Work shown above! Answer above as well

6 0

3 years ago

A,B and C are the vertices of a triangle,

Harlamova29_29 [7]

SSS (side, side, side)

3 0

2 years ago

Read 2 more answers