MATHEMATICS HIGH SCHOOL

Find the area of the blue sector. Use 3.14 for pi and round to the nearest hundredth.

Answers

Answer 1

Answer:

Answer:

The area of blue sector is 25.64 in²

Step-by-step explanation:

In the given figure, we need to find the area of blue section whose central angle is 60°

Formula:

$\text{Area of sector}= (\theta)/(360^\circ)* \pi r^2$

Radius of circle (r)=7 in

$\theta = 60^\circ$

Substitute the value into formula and find area of sector.

$\text{Area of blue sector}= (60)/(360)*3.14* 7^2$

$\text{Area of blue sector}= 25.643\approx 25.64\text{ in}^2$

Hence, The area of blue sector is 25.64 in²

Related Questions

Hich expression is rational? a. 4.121221222… b.square root of 36 c. square root of 21 d.1.192744502…
To make a specific hair dye, a hair stylist uses a ratio of 1 1/8 oz of red tone, 3/4 oz of graytone, and 5/8 oz of brown tone. If the stylist needs to make 20 oz of dye, how much ofeach dye color is needed
Answer with solutions.Find the corresponding roots in the box for the given quadratic equations and get the letters to decode the hidden message . You may use the extracting the square root methodP:±6 M: ±7 C:5,6 A : 0 J:4,-1Q:±√5 L:±11 H:±4 D:-4,1 I:±3S:16,-6 Y: ±8 B:±4√2 E:±2 A:0,-4U:6,0 U:±√10 N:6,-16 G:1,-1 T:±2√2O:±6√2 V:±√3 I:±5 J:-7,-1 K:5,-2W:±12 F:±2√3 X:±6 R:0,-6 •:9±√6/4Message : __________________________________________________________________________1. x2 = 492. x2 -27 =03. 3x2-36= 04. 9x2 = 05. 5x2- 15=06. 2x2- 144=07. ( x + 3)2 = 9 8. 4x2 -100 =09. 5x2 = 4010. 3x2 -12 = 011. (x-5)2
1/3 (n+1)+2=1/6 (3n-5)
What is the value of x in the equation x-2/3 + 1/6 =5/6 ?

7 × (–3) × (–2)2 = ?

A. 84
B. 48
C. –84
D. –48

Answers

7 × (-3) ×(-2)2 =

7 × -3 × -4 = (you have to eliminate all parenthesis in the equation.)

7 × -12 = -84

so your answer would be: C

7*-3=-21 and -2*2=-4 so -21*-4= 84
answer is 84

One positive integer is 1 greater than 3 times another positive integer. If the product of the two integers is 154, then what is the sum of the two integers?

Answers

Answer:

Step-by-step explanation:

Let the smaller integer be x.

The other integer is 1 greater than 3 times x, or 3x + 1.

The product is 154.

x(3x + 1) = 154

3x^2 + x - 154 = 0

157 = 2 * 7 * 11

14 * 11 = 154

22 * 7 = 154

(3x + 22)(x - 7) = 0

3x + 22 = 0 or x - 7 = 0

3x = -22 or x = 7

x = -22/3 or x = 7

-22/3 is not a positive integer, so we discard that solution.

The smaller integer is 7.

3x + 1 = 3(7) + 1 = 22

The greater integer is 22.

The sum of the integers is 7 + 22 = 29

Answer: 29

Step-by-step explanation:

Simplify the product using the distributive property. (-4h +2)(3h +7)

Answers

Answer: $-12h^2-22h+14$

Step-by-step explanation:

According to distributive property under multiplication over addition

We can write a.(b+c)=a.b+a.c

Since, The given expression, (-4h+2)(3h+7)

By applying distribution in first bracket.

we can write, (-4h+2)(3h+7)= -4h(3h+7)+2(3h+7)

Again on applying distribution property,

we get, (-4h+2)(3h+7)= -4h×3h+(-4h)×7+2×3h+2×7

⇒ $(-4h+2)(3h+7)= -12h^2-28h+6h+14$ = $-12h^2-22h+14$

-2(2h - 1)(3h + 7)
(-2(2h) - 2(-1))(3h + 7)
(-4h + 2)(3h + 7)
-4h(3h + 7) + 2(3h + 7)
-4h(3h) - 4h(7) + 2(3h) + 2(7)
-12h² - 28h + 6h + 14
-12h² - 22h + 14

Solve for x in the following equation: ax+5x-4=10

Answers

Factor in "x" the expression:

$ax+5x-4=10\n \n (a+5)x-4=10\n \n (a+5)x=10+4\n \n (a+5)x=14\n \n \boxed{x=(14)/(a+5)}$

Prove that: ${ \left( { e }^{ \sqrt { { e }^{ \ln { \left( \frac { { 3 }^( 0 ) }{ \sin { \left( \frac { \pi }{ 2 } \right) } } \right) } } } } \right) }^{ \ln { \left( \sqrt { { e }^{ \ln { \left( \frac { { 3 }^( 0 ) }{ \sin { \left( \frac { \pi }{ 2 } \right) } } \right) } } } \right) } }=1$

Show your workings.

Answers

${ \left( { e }^{ \sqrt { { e }^{ \ln { \left( \frac { { 3 }^( 0 ) }{ \sin { \left( \frac { \pi }{ 2 } \right) } } \right) } } } } \right) }^{ \ln { \left( \sqrt { { e }^{ \ln { \left( \frac { { 3 }^( 0 ) }{ \sin { \left( \frac { \pi }{ 2 } \right) } } \right) } } } \right) } }=1\n{ \left( { e }^{ \sqrt { { e }^{ \ln { \left( \frac { 1 }{ 1} \right) } } } } \right) }^{ \ln { \left( \sqrt { { e }^{ \ln { \left( \frac { 1 }{1 } \right) } } } \right) } }=1\n$
${ \left( { e }^{ \sqrt { { e }^( \ln 1 ) } } \right) }^{ \ln { \left( \sqrt { { e }^( \ln 1 ) } \right) } }=1\n{ \left( { e }^( \sqrt 1 ) \right) }^{ \ln { \left( \sqrt 1 \right) } }=1\n{ \left( { e }^ 1 \right) }^( \ln 1 )=1\n e ^( \ln 1 )=1\n1=1$

Given the fu8nction f(x) = 2(x + 10), find x if f(x) = 24.A) 68
B) 7
C) 17
D) 2

Answers

f(x)=2(x+10)=24
2(x+10)=24
divide 2
x+10=12
minus 10
x=2
D

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