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maxonik [38]
3 years ago
7

5. Name a point coplanar to point K?

Mathematics
1 answer:
Zielflug [23.3K]3 years ago
6 0
Points M and N are coplanar to point K. This is because they are the only other points that share the plane with Point K.
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What is the domain (in interval notation) of the following functions?
alexgriva [62]
1.\\g(x)=\frac{3}{5x-4}\\\\D:5x-4\neq0\to5x\neq4\ \ \ /:5\to x\neq\frac{4}{5}\to x\in\mathbb{R}\ \backslash\ \{\frac{4}{5}\}\\\\2.\\h(x)=\frac{\sqrt{x}}{x-5}\\\\D:x\geq0\ \wedge\ x-5\neq0\to x\geq0\ \wedge\ x\neq5\to x\in\left

3.\\f(x)=\frac{\sqrt{x}}{x^2-5x}\\\\D:x\geq0\ \wedge\ x^2-5x\neq0\to x\geq0\ \wedge\ x(x-5)\neq0\\\\\to x\geq0\ \wedge\ x\neq0\ \wedge\ x\neq5\to x\in\mathbb{R^+}\ \backslash\ \{5\}\\\\4.\\g(x)=\frac{\sqrt{x}+5}{x^2-x-20}\\\\D:x\geq0\ \wedge\ x^2-x-20\neq0\to x\geq0\ \wedge\ (x+4)(x-5)\neq0\\\\\to x\geq0\ \wedge\ x\neq-4\ \wedge\ x\neq5\to x\in\left

5.\\h(x)=\frac{3}{x^2+1}\\\\D:x^2+1\neq0\to x^2\neq-1\to x\in\mathbb{R}\\\\6.\\f(x)=\frac{\sqrt{x-2}}{x+1}\\\\D:x-2\geq0\ \wedge\ x+1\neq0\to x\geq2\ \wedge\ x\neq-1\to x\in\left

7.\\g(x)=\frac{x^2}{3x^2-x-2}\\\\D:3x^2-x-2\neq0\to (3x+2)(x-1)\neq0\to x\neq-\frac{2}{3}\ \wedge\ x\neq1\\\\\to x\in\mathbb{R}\ \backslash\ \{-\frac{2}{3};\ 1\}\\\\8.\\h(x)=3(x-4)^2-7\\\\D:x\in\mathbb{R}
4 0
3 years ago
Brock is a plumber. He charges a flat rate of $40 to visit a house to inspect it's plumbing. He charges an additional $20 for ev
Jet001 [13]

Answer:

I believe the answer is A - graph W

Step-by-step explanation:


6 0
4 years ago
What is the value of x in the number 5/3+ 3x+4=8​
amm1812
5/3+3x+4= 8
= 3x= 4-5/3
= 3x= 7/3
= x= 7/9
4 0
3 years ago
Find the horizontal and vertical asymptotes of​ f(x). ​f(x) equals = StartFraction 6 x Over x plus 2 EndFraction 6x x+2 Find the
miss Akunina [59]

Answer:

The horizontal asymptote can be described by the line y = 6

The vertical asymptote can be described by the line x = -2

Step-by-step explanation:

* <em>Lets the meaning of vertical and horizontal asymptotes</em>

- <u><em>Vertical asymptotes</em></u> are vertical lines which correspond to the zeroes

  of the denominator of a rational function

- <u><em>A horizontal asymptote</em></u> is a y-value on a graph which a function

 approaches but does not actually reach

- If the degree of the numerator is less than the degree of the

 denominator, then there is a horizontal asymptote at y = 0

- If the degree of the numerator is greater than the degree of the

 denominator, then there is no horizontal asymptote

- If the degree of the numerator is equal the degree of the denominator,

 then there is a horizontal asymptote at y = leading coefficient of the

 numerator ÷ leading coefficient of the denominator

* <em>Lets solve the problem</em>

∵ f(x)=\frac{6x}{x+2}

∵ The numerator is 6x

∵ The denominator is x + 2

∴ The numerator and the denominator have same degree

∵ The leading coefficient of the numerator is 6

∵ The leading coefficient of the denominator is 1

∴ There is a horizontal asymptote at y = 6/1

∴ <em>The horizontal asymptote can be described by the line y = 6</em>

- Put the denominator equal zero to find its zeroes

∵ The denominator is x + 2

∴ x + 2 = 0

- Subtract 2 from both sides

∴ x = -2

∴ <em>The vertical asymptote can be described by the line x = -2</em>

6 0
3 years ago
Read 2 more answers
suppose a population of 250 crickets doubles in size every six months. How many crickets will there be after two years?
lord [1]

double every 6 months 

in 6 months there would be 250*2 = 500 crickets

 in 1 year there would be 500*2 = 1000 crickets

in 1.5 years there would be 1000 *2 = 2000 crickets

 in 2 years there would be 2000 *2 = 4000 crickets

4 0
3 years ago
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