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2 years ago

Given f(x) = -x – 4, find f(4). I need help

Mathematics

1 answer:

Brums [2.3K]2 years ago

5 0

Answer:

f(4) = -8

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Algebra I</u>

Function Notation

Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = -x - 4

f(4) is x = 4

<u>Step 2: Evaluate</u>

Substitute in <em>x</em>: f(4) = -4 - 4
Subtract: f(4) = -8

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Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

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2 years ago

Lydia graphed ADEF at the coordinates D (-2,-1), E (-2, 2), and F (0,0). She thinks ADEF is a right triangle. Is Lydia's asserti

Setler79 [48]

The true statement is (c) No; the slopes of segment EF and segment DF are not opposite reciprocals.

<h3>Right triangles </h3>

Right triangles have a pair of perpendicular lines

Coordinates

The coordinates are given as:

D = (-2,-1)
E = (-2,2)
F = (0,0)

<h3>Slopes</h3>

Start by calculating the slopes of lines DF and EF using:

$m = \frac{y_2 -y_1}{x_2 -x_1}$

So, we have:

$m_{DF} = \frac{0 + 1}{0 +2}$

$m_{DF} = \frac{1}{2}$

Also, we have:

$m_{EF} = \frac{0 -2 }{0+2}$

$m_{EF} = \frac{-2 }{2}$

$m_{EF} = -1$

For the triangle to be a right triangle, then the calculated slopes must be opposite reciprocals.

i.e.

$m_1 = -\frac{1}{m_2}$

By comparison, the slopes of both lines are not opposite reciprocals.

Hence, the true statement is (c)