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2 years ago

This means that each point where the two lines intersect is a

1 answer:

8 0

Answer: When two or more lines cross each other in a plane, they are called intersecting lines. The intersecting lines share a common point, which exists on all the intersecting lines, and is called the point of intersection. Here, lines P and Q intersect at point O, which is the point of intersection.

I think this the answer‍♀️

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Help me with question 7 pls

Marrrta [24]

If the coefficient of x^2 is negative, the graph will be n shaped and curve down instead of like a u shape if it was positive. If the vertex is below the x axis and curves down, it won’t pass the x axis, Tia is right.

5 0

2 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

HELP AGAIN ITS MY LAST QUESTION

Lera25 [3.4K]

Answer:

$\frac{43}{5}, \frac{-7}{5}$

Step-by-step explanation:

6 0

2 years ago

Please I will rate you with 5 stars just please help me with this problem

olchik [2.2K]

Answer:

Step-by-step explanation:

If we isolate the variable in each given compound inequality, we can quickly realize that D is correct.

$16\geq 3x+4>1\\12\geq 3x>-3\\4\geq x>-1$

x is less than or equal to four (solid point on 4 with line going left), but greater than -1 (open point going right).

Signs with 'or equal to' have solid points, and signs without are not solid.

I hope this helps!

8 0

2 years ago

-2/5x+6=8 can you help me

ale4655 [162]

If you would like to solve the equation - 2/5 * x + 6 = 8, you can do this using the following steps:

- 2/5 * x + 6 = 8
- 2/5 * x = 8 - 6
- 2/5 * x = 2 /*(-5/2)
x = 2 * (- 5/2)
x = - 5

The correct result would be -5.

7 0

2 years ago

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