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2 years ago

What is the recirpocal of 21/32

Mathematics

1 answer:

soldier1979 [14.2K]2 years ago

3 0

The product of a number and its reciprocal would be equal to one

So just flip the numbers

=32/21

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Write 9 as a power of the base 3

malfutka [58]

9 = 3^2
Since 3 squared equals 9, that is how 9 as a power of the base 3 should be written.

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2 years ago

2852.1 rounded to the nearest tenth of a mile

katrin2010 [14]

2852.1 rounded to the nearest tenth of a mile is still 2852.1

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3 years ago

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Write an equation for a circle with center at (5,-2) and diameter of 8 centimeters.

mars1129 [50]

Answer:

$(x-5)^2+(y+2)^2=16$

Step-by-step explanation:

lol brainlessly...

General Circle function: $(x-h)^2+(y-k)^2=r^2$ , where h is the x-coordinate of the center, k is the y-coordinate of the center, and r is the radius.

The radius of the circle is half the diameter so: 8 / 2 = 4

The x-coordinate of the circle is 5

The y-coordinate of the circle is -2

Plug in.

$(x-5)^2+(y-(-2))^2=4^2$

$(x-5)^2+(y+2)^2=16$

5 0

3 years ago

Use Cramer Rule to solve the following system: 8x−5y=70 and 9x+7y=3

nlexa [21]

Answer:

$(x,y) = (5,-6)$

Step-by-step explanation:

$\underline{\textbf{Determinant of a matrix.}}\\\\\text{For a}~ 2 \times 2 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2\\b_1&b_2 \end{vmatrix} = a_1b_2 - a_2b_1\\\\\\\text{For a}~ 3 \times 3 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix} = a_1\begin{vmatrix} b_2&b_3\\c_2&c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1&b_3\\c_1&c_3 \end{vmatrix}+ a_3 \begin{vmatrix} b_1&b_2\\c_1&c_2 \end{vmatrix}\\\\\\$

$~~~~~~~~~~~~~~~~~~=a_1(b_2c_3-b_3c_2) -a_2(b_1c_3-b_3c_1) +a_3(b_1c_2-b_2c_1)$

$\underline{\textbf{Cramer's Rule to solve a system of two equations.}}\\\\\text{Consider the system of two equations:}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_1x + b_1 y= c_1\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_2x +b_2 y = c_2\\\\\text{Here,}\\\\x = \dfrac{D_x}{D}= \dfrac{\begin{vmatrix} c_1&b_1\\c_2&b_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\\\ y= \dfrac{D_y}{D}= \dfrac{\begin{vmatrix} a_1&c_1\\a_2&c_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\$

$\underline{\textbf{Solution:}}\\\\~~~~~~~~~~~~~~~~~~~~~~~8x-5y = 70~~~~~~...(i)\\\\~~~~~~~~~~~~~~~~~~~~~~~9x +7y = 3~~~~~~~...(ii)\\\\\text{Applying Cramer's rule:}\\\\x = \dfrac{D_x}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 70& -5 \\3&7 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{70(7) -(-5)(3)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{490+15}{56+45}\\\\\\~~=\dfrac{505}{101}\\\\\\~~=5$

$y = \dfrac{D_y}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 8& 70 \\9&3 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{(8)(3) -(70)(9)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{24-630}{56+45}\\\\\\~~=-\dfrac{606}{101}\\\\\\~~=-6$

$\textbf{Hence, the solution to the system of equation is}~ (x,y) = (5,-6)$

7 0

1 year ago

HELP PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!

Marrrta [24]

Answer:

For 10 plot a point 3 lines back and 5 lines back

For 11 its 2

Step-by-step explanation:

7 0

3 years ago

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