If f and t are both even functions, is f 1 t even? If f and t are both odd functions, is f 1 t odd? What if f is even and t is odd? Justify your answers.

Answers

Answer 1

Answer:

If the $f(x)$ and $t(x)$ are even function then $fo\ t\ (x)$ is an even function, if $f(x)$ and $t(x)$ are odd function then the function $fo\ t\ (x)$ is an odd function and if $f(x)$ is even and $t(x)$ is odd then the function $fo\ t\ (x)$ is an even function.

Further explanation:

An even functrion satisfies the property as shown below:

$\boxed{f(-x)=f(x)}$

An odd functrion satisfies the property as shown below:

$\boxed{f(-x)=-f(x)}$

Consider the given composite function as follows:

$\boxed{fo\ t\ (x)=f\left(t(x))\right}$

If both the function $f(x)$ and $t(x)$ are even function.

$\begin{aligned}fo\ t\ (-x)&=f\left(t(-x))\right\n&=f\left(t(x))\right\n&=fo\ t\ (x)\end{aligned}$

From the above calculation it is concluded that,

$\boxed{fo\ t\ (-x)=fo\ t\ (x)}$

This implies that the composite function $fo\ t\ (x)$ is an even function.

If both the function $f(x)$ and $t(x)$ are odd function.

$\begin{aligned}fo\ t\ (-x)&=f\left(t(-x))\right\n&=f\left(-t(x))\right\n&=-fo\ t\ (x)\end{aligned}$

From the above calculation it is concluded that,

$\boxed{fo\ t\ (-x)=-fo\ t\ (x)}$

This implies that the composite function $fo\ t\ (x)$ is an odd function.

If the function $f(x)$ is even and $t(x)$ is odd.

$\begin{aligned}fo\ t\ (-x)&=f\left(t(-x))\right\n&=f\left(-t(x))\right\n&=fo\ t\ (x)\end{aligned}$

From the above calculation it is concluded that,

$\boxed{fo\ t\ (-x)=fo\ t\ (x)}$

This implies that the composite function $fo\ t\ (x)$ is an even function.

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Answers

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Answers

This isn't an equation, so you cannot "solve for x." However, you can factor, as follows:

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What are the trigonometric ratios

Answers

The trig ratios are a set of numbers that are properties
of every angle.

When the angle is in a right triangle, they can be defined
like this:

Sine (sin) of the angle: (opposite leg) divided by (hypotenuse)
Cosine (cos) of the angle: (adjacent leg) divided by (hypotenuse)

Tangent (tan) of the angle: (opposite leg) divided by (adjacent leg)
Cotangent (cot) of the angle: (adjacent leg) divided by (opposite leg)

Secant (sec) of the angle: (hypotenuse) divided by (adjacent leg)
Cosecant of the angle: (hypotenuse) divided by (opposite leg).

The size of the right triangle doesn't matter, only the ratios of
pairs of sides. That will always be the same number for the
same angle, no matter how big or small the triangle is.

There's no easy way to calculate the numbers for an angle.
You just have to look them up, using tables in books, or online
(try 'cosine 29' on Google), or on most hand-calculators.

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Answers

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hope this helps :) it took me lots of editing hope it helpful btw my spelling is off

The minimum of the graph of a quadratic function is located at (–1, 2). The point (2, 20) is also on the parabola. Which function represents the situation?A.f(x) = (x + 1)2 + 2
B.f(x) = (x – 1)2 + 2
C.f(x) = 2(x + 1)2 + 2
D.f(x) = 2(x – 1)2 + 2

Answers

Minimum wykresu funkcji kwadratowej znajduje się w ( -1, 2). Punkt ( 2 , 20) jest również od paraboli. Która funkcja reprezentuje sytuację?
Canonical form of the function
f(x) = a* (x - p)² + q

A .f(x) = (x + 1)² + 2 ⇔ p= -1 , q = 2
B. f(x) = (x – 1)² + 2 we reject
C. f(x) = 2(x + 1)² + 2 ⇔ p = -1 , q = 2
D .f(x) = 2(x – 1)² + 2 we reject

The point (2,20) substitute
A f(x) = (x +1)² + 2
20 = (2 + 1 )² + 2
20 ≠ 9 +2
20 ≠ 11 we reject

D f(x) = 2* (x + 1)² + 2
20 = 2* (2+1)² + 2
20 = 2 * 3² + 2
20 = 2 * 9 + 2
20 = 18 + 2

Reply C

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