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3 years ago

13

in the equation 6+3y=4y+2 the variable y represents the same value. Is y= 2,3,4 or 5 the solution of this equation? Explain how

you know

2 answers:

kherson [118]3 years ago

3 0

6 + 3y = 4y +2
3y - 4y = 2 - 6
-y = -4
y = 4

hope that helps :)

alukav5142 [94]3 years ago

3 0

It's 4 because if you multiply 4 with the Y numbers. 6 + 3y= 6 + 12y= 18 4y + 2= 16y + 2= 18 Hope I'm helpful

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Which function is nonlinear?

lutik1710 [3]

Answer:

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Step-by-step explanation:

U can tell by graphing them out yourself and if u see a straight line then it linear, and if you see a curved loop then it's non- linear. but I'll do it :)

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All points on a given line have their y-coordinate as 9. What is the equation of this line? A. x + y = 9 B. x – y = 9 C. x = 9 D

Phoenix [80]

D. y = 9 because it has a complete zero slope while the y-intercept of the equation is 9, so the equation is y = 9.

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3 years ago

Read 2 more answers

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

Hoochie [10]

If you had values instead of algebraic expressions, you would find the area of the patio and subtract that from the area of the yard. Even though you don't have values, you can still find the area and subtract by expanding and simplifying:
(8x + 2)(6x + 3)
8x × 6x = 48x²
8x × 3 = 24x
2 × 6x = 12x
2 × 3 = 6
So you get 48x² + 36x + 6 as your area of the yard

(x + 5)(3x + 1)
x × 3x = 3x²
x × 1 = x
5 × 3x = 15x
5 × 1 = 5
So the area of the patio is 3x² + 16x + 5

(48x² + 36x + 6) - (3x² + 16x + 5)
48x² - 3x² = 45x²
36x - 16x = 20x
6 - 5 = 1

So your answer is D. 45x² + 20x + 1. I hope this helps!

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