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2 years ago

What set does 30 belong to

Mathematics

1 answer:

docker41 [41]2 years ago

7 0

Answer:

30 belongs to the set of real numbers, the set of rational numbers, the set of whole numbers, the set of natural numbers and the set of integers.

Step-by-step explanation:

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Which one is the right answer

Jobisdone [24]

Answer: A. f(g(x)) =x and g(f(x))=x and f and g are inverses.

Step-by-step explanation:

A function g is the inverse of function f if

f(g(x)) = g(f(x))=x

Given: f(x)= x+4 , g(x) = x-4

f(g(x)) = f(x+4) = (x-4)+4

= x-4+4 = x

i.e. f(g(x)) =x

g(f(x))= g(x-4) = (x-4)+4 =x

i.e. g(f(x))=x

Hence, f(g(x)) =x and g(f(x))=x and f and g are inverses.

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2 years ago

In Mrs. Trach’s class, 18 students prefer to do their homework before dinner. In Ms. Love-Benson's class, 20 students prefer to

deff fn [24]

Yes because in Mrs. Benson‘s class there are 20 students and in Mrs Tracy’s class there are only 18

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2 years ago

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What slope does a vertical line have

Anika [276]

The slope on a vertical line will be 0

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3 years ago

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Question 8 Find the value of "X". Round to the nearest tenth.

jekas [21]

Answer:

Step-by-step explanation:

The intersecting chord theorem is expressed as;

7 * x = 4 * 14

7x = 56

Divide both sides by 7

7x/7 = 56/7

x = 8

Hence the value of x is 8

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2 years ago

find the coordinates of the point P on the parabola y=1-x^2 with domain 0≤x≤1 that minimize the area of the triangle enclosed by

coldgirl [10]

Let point P be with coordinates $(x_0,y_0).$ Find the equation of the tangent line.

1. If $y=1-x^2,$ then $y'=-2x.$

2. The equation of the tangent line at point P is

$y-y_0=-2x_0(x-x_0).$

Find x-intercept and y-intercept of this line:

when x=0, then $y=y_0+2x_0^2;$
when y=0, then $x=\dfrac{y_0}{2x_0}+x_0=\dfrac{y_0+2x_0^2}{2x_0}.$

The area of the triangle enclosed by the tangent line at P, the x-axis, and y-axis is

$A=\dfrac{1}{2}\cdot (2x_0^2+y_0)\cdot \left(\dfrac{y_0+2x_0^2}{2x_0}\right)=\dfrac{(y_0+2x_0^2)^2}{4x_0}.$

Since point P is on the parabola, then $y_0=1-x_0^2$ and

$A=\dfrac{(1-x_0^2+2x_0^2)^2}{4x_0}=\dfrac{(1+x_0^2)^2}{4x_0}.$

Find the derivative A':

$A'=\dfrac{2(1+x_0^2)\cdot 2x_0\cdot 4x_0-4(1+x_0^2)^2}{16x_0^2}=\dfrac{12x_0^4+8x_0^2-4}{16x_0^2}.$

Equate this derivative to 0, then

$12x_0^4+8x_0^2-4=0,\\ \\3x_0^4+2x_0^2-1=0,\\ \\D=2^2-4\cdot 3\cdot (-1)=16,\ \sqrt{D}=4,\\ \\x_0^2_{1,2}=\dfrac{-2\pm4}{6}=-1,\dfrac{1}{3},\\ \\x_0^2=\dfrac{1}{3}\Rightarrow x_0_{1,2}=\pm\dfrac{1}{\sqrt{3}}.$

And

$y_0=1-\left(\pm\dfrac{1}{\sqrt{3}}\right)^2=\dfrac{2}{3}.$

Answer: two points: $P_1\left(-\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right), P_2\left(\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right).$

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2 years ago