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4 years ago

8

Circle A has a circumference of 8/3 m. Circle B has a diameter that is 3/2 times as long as Circle A’s diameter. What is the cir

cumference of Circle B?

1 answer:

patriot [66]4 years ago

6 0

Answer:

the answer is 4

Step-by-step explanation:

if my calculations are right

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What is the solution to the system of linear equations?<br><br> −9x+4y=55<br><br> −11x+7y=82

LuckyWell [14K]

Here is my process with mathematical expression interpreter, LaTeX. Full process given below to obtain the values of variable "x" and "y" altogether, solutions to these system of linear equations.

$\begin{bmatrix}-9x & + & 4y & = & 55 & \bf{- - - \: Eq. \: 1} \\ \\ -11x & + & 7y & = & 82 & \bf{- - - \: Eq. \: 2} \end{bmatrix}$

Now here, we should isolate a variable, or take it as a separate form to find the equation, and furthermore substitute the value of variable "y" into the original isolation of "x", to obtain both the solutions for this linear system of equation. Perform this on equation number 1 (Eq. 1).

Subtract the variable attached value by "4y" on both the sides, in current expression.

$\mathbf{-9x + 4y - 4y = 55 - 4y}$

$\mathbf{-9x = 55 - 4y}$

Both the sides, perform a division of value "-9".

$\mathbf{\dfrac{-9x}{-9} = \dfrac{55}{-9} - \dfrac{4y}{-9}}$

$\mathbf{x = \dfrac{55 - 4y}{-9}}$

$\mathbf{x = - \dfrac{55 - 4y}{9}}$

Substitute or just plug the value of newly obtained expression for variable "x" into Equation, numbered as "2" (Eq. 2.) and isolate further for the variable "y", to obtain first solution for this linear equation.

$\mathbf{-11 \Bigg(- \dfrac{55 - 4y}{9} \Bigg) + 7y = 82}$

$\mathbf{\dfrac{(55 - 4y) \times 11}{9} + 7y = 82}$

Multiply both the sides by a value of "9".

$\mathbf{\dfrac{(55 - 4y) \times 11}{9} + 7y \times 9 = 82 \times 9}$

$\mathbf{11 (55 - 4y) + 63y = 738}$

$\mathbf{605 - 44y + 63y = 738}$

$\mathbf{605 + 19y = 738}$

Subtract both the sides by a value of "- 605".

$\mathbf{605 + 19y - 605 = 738 - 605}$

$\mathbf{19y = 133}$

Divide both the sides by "19".

$\mathbf{\dfrac{19y}{19} = \dfrac{133}{19}}$

$\boxed{\mathbf{y = 7}}$

Substitute this variable value of "y = 7" , into our original isolation for variable "x", the expression is to be substituted by that value to complete the solutions for the linear equations. That is:

$\mathbf{x = - \dfrac{55 - 4y}{9}; \quad y = 2}$

$\mathbf{\therefore \quad x = - \dfrac{55 - 4 \times 7}{9}}$

$\mathbf{\therefore \quad x = - \dfrac{55 - 28}{9}}$

$\mathbf{x = - \dfrac{27}{9}}$

$\boxed{\mathbf{x = - 3}}$

Finalised solutions for these linear system of equations for two components , is:

$\boxed{\mathbf{\underline{\therefore \quad Final \: Solutions \: for \: these \: System \: of \: Linear \: Equations: \: x = - 3, \: \: y = 7}}}$

Hope it helps.

5 0

3 years ago

H over 3 plus 16 equals 18

3241004551 [841]

Answer:

If you're looking for H it could be 6

Step-by-step explanation:

bc 6/3+16=18 so 6/3=2 so 2+16=18

3 0

3 years ago

Read 2 more answers

I NEEEEEEEED HEEEEEEEEEEEEELLLLLLLLLLLLLLLPPPPPPPPPPPPP PPPPPPPPPPPLLLLLLLLLLLLLLLLZZZZZZZZZZZZZ

Serga [27]

Answer:

Hey! I have two answers for you. If x is next to the denominator of 7 then its −1/35 but if x is next to the whole fraction then its just -35

Step-by-step explanation:

4 0

3 years ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

4 years ago

Solve the system of equations below.<br> y=x-1<br> x+y=3

Airida [17]

Answer:

x=2, y=1

Step-by-step explanation:

y=x-1

y=3-x

x-1=3-x

2x=4

x=2

4 0

3 years ago