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2 years ago

Solve the following equation for I. A = 2 πr + πrl. Explain each step. BRAINLY IF CORRECT

Mathematics

1 answer:

hram777 [196]2 years ago

4 0

Answer:

Step-by-step explanation:

A = 2π +πrl

A - 2π = πrI subtract by 2π

(A- 2π)/(πr) = I divide by πr to isolate I

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Can someone help me please

Alborosie

Answer:

Step-by-step explanation:

Count the units (Squares) from the first point to the second and you will see that it is 7 units away.

6 0

3 years ago

Read 2 more answers

Let 3.27<<5.89 represent an interval on the number line. Find the value that is in the middle of the interval.

laila [671]

Answer:

The middle value of the interval is 4.58

Step-by-step explanation:

Consider the provided interval.

We need to find the value that is in the middle of the interval.

We can find the middle value of the interval by adding the upper and lower limit and divide the sum of the upper and lower limits by 2.

Here the upper limit is 3.27 and lower limit is 5.89.

$\frac{3.27+5.89}{2}=\frac{9.16}{2}=4.58$

Hence, the middle value of the interval is 4.58

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3 years ago

14) A circular clock has a circumference of 80.5 meters. What the measurement of the radius, in meters, of the clock? Round to t

Liula [17]

Answer: 12.8

Step-by-step explanation:

Divide 80.5 by pi then divide by 2

7 0

2 years ago

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klasskru [66]

$\bf |x^2+3x-1|\ \textless \ 3\implies 
\begin{cases}
+(x^2+3x-1)\ \textless \ 3\\\\
-(x^2+3x-1)\ \textless \ 3
\end{cases}\\\\
-------------------------------\\\\
x^2+3x-1\ \textless \ 3\implies x^2+3x-4\ \textless \ 0\implies (x+4)(x-1)\ \textless \ 0
\\\\\\
x\ \textless \ 
\begin{cases}
-4\\
1
\end{cases}\\\\
-------------------------------\\\\
-(x^2+3x-1)\ \textless \ 3\implies x^2+3x-1\boxed{\ \textgreater \ }-3
\\\\\\
x^2+3+2\ \textgreater \ 0\implies (x+2)(x+1)\ \textgreater \ 0\implies x\ \textgreater \ 
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8 0

3 years ago

Ill give you Brainlyest! I need help ASAP work shown, please!!<br> problems 8-15

8_murik_8 [283]

Using the segment addition theorem, the solutions are:

5. x = 17

6. x = 7

7. x = 14

BC = 27

CD = 61

BD = 88

8. AB = 26

9. LJ = 46

10. x = 3

11. FG = 15

12. QS = 34

13. BC = 26

14. EG = 19

15. QS = 68

<h3>How to Apply the Segment Addition Theorem?</h3>

The segment addition theorem would be used to solve the problems as shown below:

5. UW = UV + VW [segment addition theorem]

Substitute

6x - 35 = 19 + 4x - 20

6x - 4x = 35 + 19 - 20

2x = 34

x = 17

6. HJ = HI + IJ [segment addition theorem]

Substitute

7x - 27 = 3x - 5 + x - 1

7x - 3x - x = 27 - 5 - 1

3x = 21

x = 7

7. BD = BC + CD [segment addition theorem]

Substitute

7x - 10 = 4x - 29 + 5x - 9

7x - 4x - 5x = 10 - 29 - 9

-2x = -28

x = 14

BC = 4x - 29 = 4(14) - 29 = 27

CD = 5x - 9 = 5(14) - 9 = 61

BD = 7x - 10 = 7(14) - 10 = 88

8. BC = BD

Substitute

2x + 1 = 5x - 26

2x - 5x = -1 - 26

-3x = -27

x = 9

AB = 43 - BC

AB = 43 - 2x + 1 = 43 - 2(9) + 1 = 26

9. 7x - 10 = 9x - 11 - (x + 3)

7x - 10 = 9x - 11 - x - 3

7x - 9x + x = 10 - 11 - 3

-x = -4

x = 4

LJ = 28 + 7x - 10 = 28 + 7(4) - 10

LJ = 46

10. 8x + 11 = 12x - 1

8x - 12x = -11 - 1

-4x = -12

x = 3

11. 11x - 7 = 3x + 9

11x - 3x = 7 + 9

8x = 16

x = 2

FG = 11x - 7 = 11(2) - 7

FG = 15

12. 5x - 3 = 21 - x

5x + x = 21 + 3

6x = 24

x = 4

QS = 2(5x - 3) = 10x - 6 = 10(4) - 6

QS = 34

13. 8x - 20 = 2(3x - 1)

8x - 20 = 6x - 2

8x - 6x = 20 - 2

2x = 18

x = 9

BC = AB = 3x - 1 = 3(9) - 1

BC = 26

14. 5x - 1 = 7x - 13

5x - 7x = 1 - 13

-2x = -12

x = 6

EG = 6x - 4 - 13 = 6(6) - 4 - 13

EG = 19

15. RT - ST = RS

Substitute

8x - 43 - (4x - 1) = 2x - 4

8x - 43 - 4x + 1 = 2x - 4

4x - 42 = 2x - 4

4x - 2x = 42 - 4

2x = 38

x = 19

QS = 2(RS)

QS = 2(2x - 4) = 4x - 8 = 4(19) - 8

QS = 68

Learn more about the segment addition theorem on:

brainly.com/question/28263183

#SPJ1

8 0

2 years ago