Her cousin is 1/3 Cheryl's age. Find her cousin's age in terms of x.

Which functions are symmetric with respect to the origin?

Damm [24]

Answer:

The correct option is;

y = arcsinx and y = arctanx

Step-by-step explanation:

The given options are;

1) y = arcsinx and y = arccosx

Here, we have at the origin, where x = 0, arccosx ≈ 1.57 while arcsinx = 0

Therefore arccosx does not intersect arcsinx at the origin for it to be symmetrical to arcsinx or the origin

2) y = arccosxy and y = arctanx

Here arctanx = 0 when x = 0 and arcos x = 1.57 when x = 0 therefore, they are not symmetrical

3) y = arctanx and y = arccotx

Similarly, At x = 0, arccotx = 1.57 therefore, they are not symmetrical

4) y = arcsinx and y = arctanx

Both functions arcsinx and arctanx pass through the origin and their shapes are similar but inverted as they go from negative to positive.

3 0

3 years ago

which of the functions have a range of real numbers greater than or equal to 1 or less than or equal to-1

lilavasa [31]

The question is incomplete. Here is the complete question:

Which of the functions have a range of all real numbers greater than or equal to 1 or less than or equal to -1? check all that apply.

A. $y=\sec x$

B. $y= \tan x$

C. $y= \cot x$

D. $y= \csc x$

Answer:

A. $y=\sec x$

D. $y=\csc x$

Step-by-step explanation:

Given:

The range is greater than or equal to 1 or less than or equal to -1.

The given choices are:

Choice A: $y=\sec x$

We know that, the $\sec x=\frac{1}{\cos x}$

The range of $\cos x$ is from -1 to 1 given as [-1, 1]. So,

$|\cos x|\leq 1\\\textrm{Taking reciprocal, the inequality sign changes}\\\frac{1}{|\cos x|}\geq 1\\|\sec x|\geq 1$

Therefore, on removing the absolute sign, we rewrite the secant function as:

$\sec x\leq -1\ or\ \sec x\geq 1\\$

Therefore, the range of $y=\sec x$ is all real numbers greater than or equal to 1 or less than or equal to-1.

Choice B: $y= \tan x$

We know that, the range of tangent function is all real numbers. So, choice B is incorrect.

Choice C: $y= \cot x$

We know that, the range of cotangent function is all real numbers. So, choice C is incorrect.

Choice D: $y=\csc x$

We know that, the $\csc x=\frac{1}{\sin x}$

The range of $\sin x$ is from -1 to 1 given as [-1, 1]. So,

$|\sin x|\leq 1\\\textrm{Taking reciprocal, the inequality sign changes}\\\frac{1}{|\sin x|}\geq 1\\|\csc x|\geq 1$

Therefore, on removing the absolute sign, we rewrite the cosecant function as:

$\csc x\leq -1\ or\ \csc x\geq 1\\$

Therefore, the range of $y=\csc x$ is all real numbers greater than or equal to 1 or less than or equal to-1.

6 0

3 years ago

In ΔABC, if AB = 10 and BC = 6, AC can NOT be equal to

Goshia [24]

Answer:

16

Step-by-step explanation:

adding any two sides of a triangle is allways greater than the third side

6 0

3 years ago

-0.125 as a fraction

kaheart [24]

Answer:

-1/8

Step-by-step explanation.

125 x 8 = 1000

.125 x 8 = 1

-.125 x 8 = -1

-1/8

8 0

3 years ago

Simplify 3/4(1/2×-12)+4/5

Wittaler [7]

Answer: $\frac{3}{8}x-\frac{41}{5}$ or $\frac{3}{8}x-8\ \frac{1}{5}$

Step-by-step explanation:

Given the following expression:

$\frac{3}{4}(\frac{1}{2}x-12)+\frac{4}{5}$

You can follow these steps in order to simplify it:

1. You need to apply Distributive property:

$=\frac{3}{8}x-9+\frac{4}{5}$

2. And now you must add the like terms. Then:

$=\frac{3}{8}x-\frac{41}{5}$

Therefore you get the following answer:

$\frac{3}{8}x-\frac{41}{5}$ or $\frac{3}{8}x-8\ \frac{1}{5}$

7 0

3 years ago