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3 years ago

In a number line when is the circle open and when is it closed?

Mathematics

1 answer:

EleoNora [17]3 years ago

7 0

The circle is closed when it is = and open when it is not = to the number that is being circled.

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Pleasee help me with this anyone..

vlabodo [156]

Answer:

Step-by-step explanation:

You need to set up a proportion

Let x = NK

7/13 = x/56 Notice that the longest side of the small trapezoid is the denominator of the fraction on the left. That means that the longest side of the large trapezoid must also be the denominator of that fraction on the right.

Cross multiply

13x = 7*56 Combine the right

13x = 392 Divide by 13

x = 392/13

x = 30.15

NK = 30.15

3 0

2 years ago

Read 2 more answers

Find the difference.

joja [24]

7x - (25x^2 + 12x) = 7x - 25x^2 - 12x = -25x^2 - 5x

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3 years ago

Where common multiples between three and five and one and 50

Arlecino [84]

300 is a multiple of 3,5,1, and 50

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3 years ago

Evaluate the expression using the given value 3n+7,n=12

balandron [24]

3x12=36
36+7= 43

The answer is 43

6 0

1 year ago

Find the zeros of y = x2 – 6x – 4 by completing the square.

katrin [286]

Answer:

The solutions to the quadratic equations are:

$x=\sqrt{13}+3,\:x=-\sqrt{13}+3$

Step-by-step explanation:

Given the function

$y\:=\:x^2\:-\:6x\:-\:4$

substitute y = 0 in the equation to determine the zeros

$0\:=\:x^2\:-\:6x\:-\:4$

Switch sides

$x^2-6x-4=0$

Add 4 to both sides

$x^2-6x-4+4=0+4$

Simplify

$x^2-6x=4$

Rewrite in the form (x+a)² = b

But, in order to rewrite in the form x²+2ax+a²

Solve for 'a'

2ax = -6x

a = -3

so add a² = (-3)² to both sides

$x^2-6x+\left(-3\right)^2=4+\left(-3\right)^2$

$x^2-6x+\left(-3\right)^2=13$

Apply perfect square formula: (a-b)² = a²-2ab+b²

$\left(x-3\right)^2=13$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solve

$x-3=\sqrt{13}$

Add 3 to both sides

$x-3+3=\sqrt{13}+3$

Simplify

$x=\sqrt{13}+3$

now solving

$x-3=-\sqrt{13}$

Add 3 to both sides

$x-3+3=-\sqrt{13}+3$

Simplify

$x=-\sqrt{13}+3$

Thus, the solutions to the quadratic equations are:

$x=\sqrt{13}+3,\:x=-\sqrt{13}+3$

4 0

3 years ago