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3 years ago

52:44

Mathematics

1 answer:

bagirrra123 [75]3 years ago

8 0

Answer:

Step-by-step explanation:

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Who can answer this ??<br> Win BRAINLEST !<br><br> HELP <br> Thank You In Advance !

kirill [66]

ANSWER

$( {x - 11)}^{2} + ( {y - 5)}^{2} = 16$

EXPLANATION

The given circle has center:

(11,5)

and

radius: 4

The equation of a circle with center (h,k) and radius , r is given in standard form as;

$( {x - h)}^{2} + ( {y - k)}^{2} = {r}^{2}$

We substitute the values into the formula to obtain,

$( {x - 11)}^{2} + ( {y - 5)}^{2} = {4}^{2}$

$( {x - 11)}^{2} + ( {y - 5)}^{2} = 16$

5 0

2 years ago

1.2983 rounded to nearest hundredth

exis [7]

Answer:

1.3000

Step-by-step explanation:

The 9 makes the 2 move to 3. Hope this helps.

3 0

3 years ago

How to solve the rational equation?<br><br> 1/3 + 1/2 = 1/x

tatyana61 [14]

Answer:

x = ⁶/₅

Step-by-step explanation:

$\begin{array}{rcll}\dfrac{1}{3} + \dfrac{1}{2} & = &\dfrac{1}{x} & \\\\2x + 3x & = & 6 & \text{Multiplied all terms by LCM of denominators (6x)}\\5x & = & 6 & \text{Simplified left-hand side}\\x & = & \mathbf{\dfrac{6}{5}} & \text{Divided each side by 5}\\\end{array}$

8 0

3 years ago

Compare the process of solving |x – 1| + 1 < 15 to that of solving |x – 1| + 1 > 15. Check all of the following you includ

zepelin [54]

The required Comparison of the inequalities are

The |x – 1| + 1 > 15 represents the value of x lies between 13<x<15.

The range of values encompassing the region's junction is (-13, 15).

If x is more than or equal to 15, then x-11+1>15 indicates the value of x is greater than or equal to 13. None of the regions in the intersection are empty.

<h3>What is inequality?</h3>

When comparing two numbers, an inequality indicates whether one is less than, larger than, or not equal to the other.

We take into account the various variables of the inequality

|x – 1| + 1 > 15

Therefore

|-x-1|+1-1<15-1

|-x-1|-1 <14

13<x<15

The required region lies between the inequality -13 <x< 15.

Simplify the inequality Ix-11+1 > 15 we get,

|x-1|+1 > 15

|x+1| +1-1 >15-1

|x-1| > 14

x> 15

x<-13

If x has a value between -13 and x + 15, then the expression "|x-1|+1+115" is true. The range "(-13, 15)" contains the intersection of the region.

If "|x-1|+1>15" then either "x >15" or "x-13" applies to the value of x. This region's intersection is unoccupied.