given a rectangle of length a and width b if A and B are both rational numbers would the perimeter be irrational or rational? Give an example choosing your own values for a and b

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Answer 1

Answer: The perimeter would be rational due to both the width and length being rational numbers. A rational number is a number that can be written as a fraction such as 8 it can be written as 8/1, but a irrational would be a decimal number which in this case there would be none because both length and width is rational number. For example, Lets say the width is 8 and the length is 10. once both are written into fraction form such as 8/1 and 10/1 we add them to get a perimeter of 18.
Hope this helps!

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What is the domain (in interval notation) of the following functions?1. g(x)=3/(5x-4)
2. h(x)=√(x)/(x-5)
3. f(x)=√(x)/(x^2-5x)
4. g(x)=(√(x)+5)/(x^2-x-20)
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6. f(x)=(√(x-2))/(x+1)
7. g(x)= x^2/(3x^2-x-2
8. h(x)=3(x-4)^2-7
Number sets in parenthesis are either on top of or beneath the fraction bar and ^2 here represents a number squared.

Answers

$1.\ng(x)=(3)/(5x-4)\n\nD:5x-4\neq0\to5x\neq4\ \ \ /:5\to x\neq(4)/(5)\to x\in\mathbb{R}\ \backslash\ \{(4)/(5)\}\n\n2.\nh(x)=(√(x))/(x-5)\n\nD:x\geq0\ \wedge\ x-5\neq0\to x\geq0\ \wedge\ x\neq5\to x\in\left<0;\ \infty\right)\ \backslash\ \{5\}$

$3.\nf(x)=(√(x))/(x^2-5x)\n\nD:x\geq0\ \wedge\ x^2-5x\neq0\to x\geq0\ \wedge\ x(x-5)\neq0\n\n\to x\geq0\ \wedge\ x\neq0\ \wedge\ x\neq5\to x\in\mathbb{R^+}\ \backslash\ \{5\}\n\n4.\ng(x)=(√(x)+5)/(x^2-x-20)\n\nD:x\geq0\ \wedge\ x^2-x-20\neq0\to x\geq0\ \wedge\ (x+4)(x-5)\neq0\n\n\to x\geq0\ \wedge\ x\neq-4\ \wedge\ x\neq5\to x\in\left<0;\ \infty\right)\ \backslash\ \{-4;\ 5\}$

$5.\nh(x)=(3)/(x^2+1)\n\nD:x^2+1\neq0\to x^2\neq-1\to x\in\mathbb{R}\n\n6.\nf(x)=(√(x-2))/(x+1)\n\nD:x-2\geq0\ \wedge\ x+1\neq0\to x\geq2\ \wedge\ x\neq-1\to x\in\left<2;\ \infty\right)$

$7.\ng(x)=(x^2)/(3x^2-x-2)\n\nD:3x^2-x-2\neq0\to (3x+2)(x-1)\neq0\to x\neq-(2)/(3)\ \wedge\ x\neq1\n\n\to x\in\mathbb{R}\ \backslash\ \{-(2)/(3);\ 1\}\n\n8.\nh(x)=3(x-4)^2-7\n\nD:x\in\mathbb{R}$

The line y=mx is tangent to the curve y^2 = X-C. Show that c = 1/ (4m^2)

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Hello,
Let (a,b) the intersection of the tangent and the curve.
That point must be double.

So
b=m*a
b²=a-c
==>(ma)²-a+c=0
==>m²a²-a+c=0 roots must be equal
Δ=(-1)²-4*m²c=0 => 4m²c=1==>c=1/(4m²)

Someone plz help me :(

Answers

Answer:

i can JUST GET ON YOUR CROME BOOK

Step-by-step explanation:

You suppose to fence the football ground. The fence should be two meters from the ground. Calculate the length of the fence.

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Answer:

choose the figure that model 8÷4/3

there are 94 suitcases in number one cargo bin. there are two cargo bins to be loaded on the airplane. the second bin contains 214 suitcases and the third 103 suitcases. if all three of these cargo bins are loaded onto the 747 airplane,how many total suitcases will be in the cargo area?

Answers

isnt it simple addition? i'm aint sure because i'm french but if i understand well, it suppose to be 411

Final answer:

The total suitcases in all three cargo bins, when added up (94 + 214 + 103), equals 411. Consequently, there will be 411 suitcases in the cargo area of the 747 airplane.

Explanation:

Total suitcase calculation involves simply adding up the numbers from all three cargo bins. The first bin contains

94 suitcases, the second bin 214 suitcases, and the third contains 103 suitcases. Adding all these up gives a total suitcase number as follows: 94 + 214 + 103 = 411. Therefore, if all these cargo bins are loaded onto the 747 airplane, here will be 411 suitcases in the cargo area.

Learn more about Addition here:

brainly.com/question/40265821

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The cross section of a cylinder taken parallel to the base produces which 2-dimensional shape?

Answers

A cylinder is a three-dimensional shape that has a base of a circle. Therefore, the cross-section of a cylinder taken parallel to the base is a circle. The volume of a cylinder is equal to the area of the base times the height of the shape.The area of the base is equal to the area of the circle.

A circle ircle because when you cut the cylinder, it is the shape you see.

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