Given circle and circle with radii of 6cm and 4cm respectively.




Compare the length of EF and GH.




The length of EF is ____the length of GH



The length of EF is___



The length of GH is ___

Answers

Answer 1

Solution :

Given that :

The radius of circle A = 6 cm

The radius of circle C = 4 cm

In circle A

[tex]\angle EAF = \theta = 140^\circ[/tex]

The length of arc EF = [tex]$2 \pi r \times \frac{\theta}{360^\circ}$[/tex]

                                   [tex]$=2 \times 3.14 \times 6 \times \frac{140^\circ}{360^\circ}$[/tex]

                                    = 14.653 cm

In circle C

[tex]\angle GCH = \theta = 140^\circ[/tex]

The length of arc GH = [tex]$2 \pi r \times \frac{\theta}{360^\circ}$[/tex]

                                   [tex]$=2 \times 3.14 \times 4 \times \frac{140^\circ}{360^\circ}$[/tex]

                                    = 9.769 cm    

Therefore,

The length of EF is 14.653 cm

The length of GH is 9.769 cm

The length of EF is  1.5 times the length of GH

i.e.                   14.653  = 1.5 x 9.769

                       14.653 = 14.653

Hence proved.

                             

Given Circle And Circle With Radii Of 6cm And 4cm Respectively. Compare The Length Of EF And GH.The Length

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Answer:

the surface area of cylinder is 75.36 inches²

hope it helps

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Answer:

SA = 75.36

Step-by-step explanation:

So, you use the formula and replace the letters with the values.

SA = 2*3.14*2^2+2*3.14*2*4

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SA = 75.36

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Answers

Answer:

[tex]y=\frac{3}{2}x+(-14)[/tex]

Step-by-step explanation:

Equation of a line has been given as,

[tex]y=-\frac{2}{3}x[/tex] --------(1)

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Wouldn’t it be 120?

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Step-by-step explanation:

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Answer:

The length of AB is approximately 6.07 m

Step-by-step explanation:

The given dimensions of the triangle are;

The value of x = 7.7 m

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The length of side AB = Required

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AC = x = The hypotenuse (opposite to right angle) side

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Answers

Answer:

q = 2p - r - s

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right  

Distributive Property

Equality Properties

Multiplication Property of Equality Division Property of Equality Addition Property of Equality Subtraction Property of Equality

Algebra I

Terms/Coefficients

Step-by-step explanation:

Step 1: Define

Identify

2p - q = r + s

Step 2: Solve for q

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Answers

Answer:

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Step-by-step explanation:

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Help pls!! this is a presentation I’m doing in class and I NEED this completed rn, ASAP!!! . What I need is just the answers for the picture below and I NEED the explanation to all of them, and if you can pls try to give a similar example to these and solve it or what did u use to solve it. (Only 1 example that is similar) Thank you so much!

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Answers

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Step-by-step explanation:

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Answer:

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Answers

Answer:

option E, C

Step-by-step explanation:

From the graph we will find the equation of g(x).

g(x) is a parabola with vertex ( h, k) = ( 0, 9)

Standard equation of parabola is ,  y = a (x - h)² + k

                                                     y = a (x - 0)² + 9

                                                     y = ax² + 9 ----------  ( 1 )

Now we have to find a .

To find a we will take another point through which the parabola passes .

Let it be ( 3, 0).

Substitute ( 3 , 0 ) in  ( 1 ) =>  0 = a (3 )² + 9

                                     => - 9 = 9a

                                     => a = - 1

Substitute a = - 1 in ( 1 ) => y = -1 x² + 9

                                  => y = - x² + 9

Therefore , g(x) = -x² + 9

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 [tex]h(x) = 4^{x}[/tex]

So g(x) = -x² + 9 and [tex]h(x) = 4^{x}[/tex]

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               [tex]\lim_{x \to \infty} g(x) = \lim_{x \to \infty} -x^2 + 9[/tex]

                                  [tex]= - \lim_{x\to \infty} x^2 + \lim_{x \to \infty} 9\\\\= - \infty + 9\\\\=- \infty[/tex]

g(x) decreases on ( 0 , ∞)

             [tex]\lim_{x\to \infty} h(x) = \lim_{x \to \infty} 4^{x}[/tex]

                                [tex]= \infty[/tex]

h(x) increases on ( 0, ∞)

option B : g(x) increasing on (- ∞, 0) - False

      g(x) = -x² + 9

       g( -2 ) = - (-2)² + 9

                 = - 4 + 9 = 5

       g ( -5) = - ( -5)² + 9

                =  - 25 + 9 = - 14

   As the value of x moves towards - ∞ , g(x) moves towards - ∞

   Therefore g(x) decreases on  (- ∞, 0)

Option C: y intercept of g(x) is greater than h(x) - True

         y intercept of g(x) = ( 0 , 9 )

         y intercept of h(x)  = ( 0 , 1 )

Option D : h(x) is a linear function - False

Option E : g(2) < h(2) - True

        g(x) =  -x² + 9

        g(2) = -(2)² + 9 = - 4 + 9 = 5

        h(x) = 4ˣ          

        h(2) = 4² = 16      

                           

Plz solve this question
Show the steps you need to take to solve it.

Answers

Answer:

V = 378y³

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Algebra I

Terms/Coefficients

Geometry

Volume of a Rectangular Prism: V = lwh

l is lengthw is widthh is height

Step-by-step explanation:

Step 1: Define

Identify variables

l = 9y

w = 7y

h = 6y

Step 2: Find Volume

Substitute in variables [Volume of a Rectangular Prism]:                              V = (9y)(7y)(6y)Multiply:                                                                                                             V = (63y²)(6y)Multiply:                                                                                                             V = 378y³

___________________________________

Problem:

What is the volume of this rectangular prism?

Formula for volume (v):

[tex]\quad\quad\quad\quad \boxed{\tt{v = lwh}}[/tex]

Given that:

[tex]\quad\quad\quad\quad\tt{length (l)= 9y}[/tex]

[tex]\quad\quad\quad\quad\tt{width(w)= 7y}[/tex]

[tex]\quad\quad\quad\quad\tt{height(h)= 6y}[/tex]

Solution:

[tex]\quad\quad\quad\quad\tt{v = lwh}[/tex]

[tex]\quad\quad\quad\quad\tt{v = (9y)(7y)(6y)}[/tex]

[tex]\quad\quad\quad\quad\tt{v = (9y)(42 {y}^{2}) }[/tex]

[tex]\quad\quad\quad\quad \underline{\tt{v = 378 {y}^{3} }}[/tex]

Hence, the final answer is:

[tex]\quad\quad\quad\quad \underline { \boxed{\tt{ \color{magenta}v = 378 {y}^{3} }}}[/tex]

___________________________________

#CarryOnLearning

✍︎ C.Rose❀

(when u solved it, can u show me how you got the answer because my teacher wants us the show the work as well)

ty

Answers

Answer 243mm^2

Explanation:
I first split the shape into two smaller shapes. Then I found those two areas and added them! (18x12)+(9x3)=243
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