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Sonja [21]
3 years ago
12

5 1⁄5 decimeters = centimeters

Mathematics
1 answer:
blsea [12.9K]3 years ago
5 0

Answer:

\large\boxed{5\dfrac{1}{5}\ decimeters=52\ centimeters}

Step-by-step explanation:

1\ dm=10\ cm\qquad\text{(1 decimeters = 10 centimeters)}\\\\\text{Therefore}\\\\5\dfrac{1}{5}\ dm=\dfrac{5\cdot5+1}{5}\ dm=\dfrac{26}{5}\ dm=\dfrac{26}{5}\cdot10\ cm=\dfrac{260}{5}\ cm=52\ cm

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Solve for x:a(a²+b²)x²+b²x-a​
m_a_m_a [10]

Answer:

x = a/(a² + b²) or x = -1/a  

Step-by-step explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} =\dfrac{-b\pm\sqrt{D}}{2a}

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b²  = (b² + 2a²)²

2. Solve for x

\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}

\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}

5 0
3 years ago
Use matrix addition to solve this equation: B + 15 −7 4 0 1 2 = 1 2 12 4 0 2 b11 = b12 = b13 = b21 = 4 b22 = −1 b23 = 0
Arada [10]

Answer:

b_{11}=-14,b_{12}=9,b_{13}=8,b_{21}=4,b_{22}=-1,b_{23}=0

Step-by-step explanation:

The given matrix addition is

B+\begin{bmatrix}15&-7&4\\ 0&1&2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}

We need to find the elements of matrix B.

Let B=\begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\end{bmatrix}

Substitute the value of matrix.

\begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\end{bmatrix}+\begin{bmatrix}15&-7&4\\ 0&1&2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}

After addition of two matrix we get

\begin{bmatrix}b_{11}+15&b_{12}-7&b_{13}+4\\ b_{21}+0&b_{22}+1&b_{23}+2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}

On equating both sides.

b_{11}+15=1\Rightarrow b_{11}=-14

b_{12}-7=2\Rightarrow b_{12}=9

b_{13}+4=12\Rightarrow b_{13}=8

b_{21}+0=4\Rightarrow b_{21}=4

b_{22}+1=0\Rightarrow b_{22}=-1

b_{23}+2=2\Rightarrow b_{23}=0

Therefore, the elements of matrix B are b_{11}=-14,b_{12}=9,b_{13}=8,b_{21}=4,b_{22}=-1,b_{23}=0.

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3 years ago
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Answer:

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