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3 years ago

If the area of a circle is found using A=πr2, find r if the area is 25in2 and using pi equals 3.14

Mathematics

1 answer:

Olegator [25]3 years ago

5 0

Answer:

r = 2.8 in ( to 1 dec. place )

Step-by-step explanation:

substitute the given values into the formula

25 = 3.14r² ( divide both sides by 3.14 )

r² = $\frac{25}{3.14}$ ( take the square root of both sides )

r = $\sqrt{\frac{25}{3.14} }$ ≈ 2.8 in

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Angle Addition Crossword

Elena-2011 [213]

Answer:

Step-by-step explanation:

1). m∠AEC = m∠AEB + m∠BEC

= 21° + 37°

= 58°

2). m∠BED = m∠BEC + m∠CED

= 37° + 44°

= 81°

3). m∠IKF = m∠IKH + m∠HKG + m∠GKF

= m∠IKH + m∠HKG + m∠IKH [Since, ∠IKH ≅ ∠GKF]

= 2∠IKH + m∠HKG

103° = 2∠IKH + 41°

2(∠IKH) = 103 - 41

m(∠IKH) = 31°

4). m∠AED = m∠AEB + m∠BEC + m∠CED

= 21° + 37° + 44°

= 102°

5). m∠JKG = 108°

m∠JKG = m∠JKI + m∠IKH + m∠HKG

108° = m∠JKI + 31° + 41°

m∠JKI = 108° - 72°

m∠JKI = 36°

6). m∠HKF = m∠GKF + m∠HKG

= m∠IKH + m∠HKG [Since, m∠GKF = m∠IKH]

= 31° + 41°

= 72°

7). m∠NQO = m∠MQN = 64°

8). m∠JKF = m∠JKI + m∠IKF

= 36° + 103°

= 139°

8). m∠MQO = 2(m∠NQO)

= 2(64)°

= 128°

9). m∠LQO = 156°

m∠LQM = m∠LQO - m∠MQO

= 156° - 128°

= 28°

10. m∠NQP = m∠NQO + m∠OQP

= 64° + m∠LQM [Since ∠OQP ≅ ∠LQM]

= 64° + 28°

= 92°

5 0

3 years ago

Help me with this pls

kobusy [5.1K]

Answer:

C.)-5

Step-by-step explanation:

y2-y1/x2-x1

(-9,6) (-6,-9)

-9-6=-15

-6-(-9)=3

-15/3=-5

4 0

3 years ago

Read 2 more answers

5. Find the value(s) of x so that the line containing the points (2x + 3, x + 2) and (0, 2) is

Dvinal [7]

Answer:

x = -2 or -9

Step-by-step explanation:

You want the values of x such that the line defined by the two points (2x+3, x+2) and (0, 2) is perpendicular to the line defined by the two points (x+2, -3-3x) and (8, -1).

<h3>Slope</h3>

The slope of a line is given by the slope formula:

m = (y2 -y1)/(x2 -x1)

Using the formula, the slopes of the two lines are ...

m1 = (2 -(x+2))/(0 -(2x+3)) = (-x)/(-2x-3) = x/(2x +3)

and

m2 = (-1 -(-3-3x))/(8 -(x+2)) = (2+3x)/(6 -x)

<h3>Perpendicular lines</h3>

The slopes of perpendicular lines have product of -1:

$\dfrac{x}{2x+3}\cdot\dfrac{2+3x}{6-x}=-1\\\\x(3x+2)=(2x+3)(x-6)\qquad\text{multiply by $(2x+3)(6-x)$}\\\\3x^2+2x=2x^2-9x-18\qquad\text{eliminate parentheses}\\\\x^2+11x+18=0\qquad\text{put in standard form}\\\\(x+2)(x+9)=0\qquad\text{factor}$

<h3>Solutions</h3>

The values of x that satisfy this equation are x = -2 and x = -9. The attached graphs show the lines for each of these cases.

4 0

1 year ago

which equation is in slope intercept form and represents a line with slope 3 through the point (9,-4)?

faust18 [17]

Answer:

y = 3x - 23

Step-by-step explanation:

y + 4 = 3(x - 9)

y + 4 = 3x - 27

y = 3x - 23

4 0

3 years ago

Paul has an eight-year loan with a principal of $26,900. The loan has an interest rate of 8.18%, compounded quarterly. If Paul p

babymother [125]

Answer:

The answer is "option b"

Step-by-step explanation:

Amount = $26,900

r= 0.0818 %

$i=\frac{r}{4}\\\\$

$=\frac{0.0818}{4}\\\\=0.02045$

t= 8 years

$n= 8 \times 4$

$=32$

$D=\frac{1-(1+i)^{-n}}{i}\\$

$=\frac{1-(1+0.02045)^{-32}}{0.02045}\\\\=\frac{1-(1.02045)^{-32}}{0.02045}\\\\=\frac{1-0.52319627}{0.02045}\\\\=\frac{0.47680373}{0.02045}\\\\=23.3155858$