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krek1111 [17]
3 years ago
14

Each point on the edge of a circle is equidistant from the center of the circle. The center of a circle is located at (6,3). Whi

ch point on the y-axis could be on the edge of the circle of the distance from the center of the circle to the edge is 10
Mathematics
2 answers:
denis-greek [22]3 years ago
6 0

Answer:

(0, –5)

Step-by-step explanation:

If you don't see this option there is another option, (0 , 11) is also correct

photoshop1234 [79]3 years ago
3 0

Good evening

Answer:

<h2>(0 , 5) and (0 , 11)</h2>

Step-by-step explanation:

Suppose M(x , y) is a point of the circle of center (6,3) and radius 10 and M is also a point of the y-axis then the coordonates of M should verify these two equations at the same time:

\left \{ {(x-6)^{2} +(y-3)^2=10^{2} } \atop {x=0}} \right.

Then

(0-6)² + (y-3)² = 100

then

36 + (y-3)² = 100

then

(y-3)² = 100 - 36

        = 64

then

|y-3| = 8

then

y - 3 = 8 or y - 3 = -8

then

y = 11   or  y = 5

then the only points that are on the edge of the circle and on y-axis are:

(0 , 5) and (0 , 11).

:)

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Sauron [17]

Answer: $11.06

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2 years ago
The coordinates of trapezoid ABCD are A(−4, 3), B(2, 3), C(4, −1) and D(−4, −1). A line segment runs through the trapezoid with
madam [21]
The distance from point Y to the y-axis is 4 units and the distance from point Z to the y-axis is 3 units, then the lelgth of the segment YZ is 4+3=7 units.
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5 0
3 years ago
Read 2 more answers
Find the sum of 12a + 7b, −6a − 9, and 14a − 12b.
Ivan
The answer is:  [A]:  " 20a − 5b − 9 " .
______________________________________
Explanation:
_______________________________________________
  (12a <span>+ 7b) + (−6a − 9) + (14a − 12b) =
 
   12a </span><span>+ 7b  +  1(−6a − 9) + 1(14a − 12b) =
</span>    
   12a + 7b + (1*-6a) + (1*-9) + (1*14a) + (1* -12b) =
      
    12a + 7b − 6a − 9 + 14a − 12b =  ?

Combine the "like terms:

12a − 6a  + 14a = 20a ; 

7b − 12b =  - 5b ;

and then we have "-9" ;
_____________________________________________________
So, write as:   " 20a − 5b − 9 " ;  which is:  Answer choice:  [A].
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8 0
3 years ago
Read 2 more answers
1 point
photoshop1234 [79]

Answer:

 b

Step-by-step explanation:

done it

5 0
3 years ago
Find the area of the region enclosed by the graphs of the functions
Vaselesa [24]

Answer:

\displaystyle A = \frac{8}{21}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Algebra I</u>

  • Terms/Coefficients
  • Functions
  • Function Notation
  • Graphing
  • Solving systems of equations

<u>Calculus</u>

Area - Integrals

Integration Rule [Reverse Power Rule]:                                                                 \displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

Integration Rule [Fundamental Theorem of Calculus 1]:                                      \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)

Integration Property [Addition/Subtraction]:                                                          \displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx

Area of a Region Formula:                                                                                     \displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx

Step-by-step explanation:

*Note:

<em>Remember that for the Area of a Region, it is top function minus bottom function.</em>

<u />

<u>Step 1: Define</u>

f(x) = x²

g(x) = x⁶

Bounded (Partitioned) by x-axis

<u>Step 2: Identify Bounds of Integration</u>

<em>Find where the functions intersect (x-values) to determine the bounds of integration.</em>

Simply graph the functions to see where the functions intersect (See Graph Attachment).

Interval: [-1, 1]

Lower bound: -1

Upper Bound: 1

<u>Step 3: Find Area of Region</u>

<em>Integration</em>

  1. Substitute in variables [Area of a Region Formula]:                                     \displaystyle A = \int\limits^1_{-1} {[x^2 - x^6]} \, dx
  2. [Area] Rewrite [Integration Property - Subtraction]:                                     \displaystyle A = \int\limits^1_{-1} {x^2} \, dx - \int\limits^1_{-1} {x^6} \, dx
  3. [Area] Integrate [Integration Rule - Reverse Power Rule]:                           \displaystyle A = \frac{x^3}{3} \bigg| \limit^1_{-1} - \frac{x^7}{7} \bigg| \limit^1_{-1}
  4. [Area] Evaluate [Integration Rule - FTC 1]:                                                    \displaystyle A = \frac{2}{3} - \frac{2}{7}
  5. [Area] Subtract:                                                                                               \displaystyle A = \frac{8}{21}

Topic: AP Calculus AB/BC (Calculus I/II)  

Unit: Area Under the Curve - Area of a Region (Integration)  

Book: College Calculus 10e

6 0
3 years ago
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