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lisov135 [29]
3 years ago
13

Domain and range of y = x2 + 9

Mathematics
2 answers:
andrew-mc [135]2 years ago
7 0

Answer: Find the domain by finding where the equation is defined. The range is the set of values that correspond with the domain.

Domain:  

(−∞,∞),{x|x∈R}

Range:

[9,∞),{y|y≥9}

olganol [36]2 years ago
3 0

Algebra Examples

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined. The range is the set of all valid y values.

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What is the unit rate of -4y = 18x
Wittaler [7]

Answer:

Step-by-step explanation:

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3 years ago
Solve the inequality 2x>30+5/4x
insens350 [35]

Answer:

Step-by-step explanation:

2x > 30+\frac{5}{4x} \\2x-\frac{5}{4x} > 30\\\frac{8x^2-5}{4x} > 30\\case~1\\if~x > 0\\8x^2-5 > 120x\\8x^2-120x > 5\\x^2-15x > \frac{5}{8} \\adding~(-\frac{15}{2} )^2~to~both~sides\\(x-\frac{15}{2} )^2 > \frac{5}{8}+\frac{225}{4} \\(x-\frac{15}{2} )^2 > \frac{455}{8} \\x-\frac{15}{2} < -\sqrt{\frac{455}{8} }  \\x < \frac{15}{2}-\sqrt{\frac{455}{8} } \\or~x < 0\\rejected~as~x > 0

x-\frac{15}{2} > \sqrt{\frac{455}{8} } \\x > \frac{15}{2} +\sqrt{\frac{455}{8} }

case~2

if~x < 0\\8x^2-5 < 120x\\8x^2-120x < 5\\x^2-15x < \frac{5}{8} \\adding~(-\frac{15}{2} )^2\\(x-\frac{15}{2} )^2 < \frac{5}{8} +(-\frac{15}{2} )^2\\|x-\frac{15}{2} | < \frac{5+450}{8} \\-\sqrt{\frac{455}{8} } < x-\frac{15}{2} < \sqrt{\frac{455}{8} } \\\frac{15}{2} -\sqrt{\frac{455}{8} } < x < \frac{15}{2} +\sqrt{\frac{455}{8} } \\but~x < 0\\7.5-\sqrt{\frac{455}{8} } < x < 0

8 0
2 years ago
PLEASE HELP<br>Is the function linear or nonlinear y=1/x ?<br><br> A. linear B. nonlinear
Rina8888 [55]
It's nonlinear.
..............
8 0
3 years ago
Read 2 more answers
Help me out plzz and I'll do the same
elixir [45]
I think it's -2 to -1 is the right answer maybe

but i hope it helps u !!
6 0
3 years ago
Use graphing to find the solutions to the system of equations.
Galina-37 [17]
In order to utilize the graph, first you have to distinguish which graph accurately pertains to the two functions.

This can be done by rewriting the equations in the form y = mx + b which can be graphed with ease; where m is the slope and b is the y intercept.

-x^2 + y = 1
y = x^2 + 1

So this will be a basic y = x^2 parabola where the center intercepts on the y axis at (0, 1)

-x + y = 2
y = x +2

So this will be a basic y = x linear where the y intercept is on the y axis at (0, 2)

The choice which depicts these two graphs correctly is the first choice. The method to find the solutions to the system of equations by using the graph is by determining the x coordinate of the points where the two graphed equations intersect.
3 0
3 years ago
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