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2 years ago

The temperature in Celsius would be,

Mathematics

1 answer:

elena-14-01-66 [18.8K]2 years ago

8 0

First derisive the equation for C

$\\ \sf\longmapsto F=\dfrac{9}{5}C+32$

$\\ \sf\longmapsto C+32=\dfrac{5}{9}F$

$\\ \sf\longmapsto C=\dfrac{5}{9}F-32$

F=101F

$\\ \sf\longmapsto C=\dfrac{5}{9}101-32$

$\\ \sf\longmapsto C=\dfrac{505}{9}-32$

$\\ \sf\longmapsto C=56.1-32$

$\\ \sf\longmapsto C=24.1°C$

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I hope this helps you

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3 years ago

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The number of hours he worked in all (written in fraction form) is $8\frac{1}{6} \ hours$

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2 years ago

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3 years ago

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Matrices A and B are square matrices of the same size. Prove Tr(c(A + B)) = C (Tr(A) + Tr(B)).

alexira [117]

Answer with Step-by-step explanation:

We are given that two matrices A and B are square matrices of the same size.

We have to prove that

Tr(C(A+B)=C(Tr(A)+Tr(B))

Where C is constant

We know that tr A=Sum of diagonal elements of A

Therefore,

Tr(A)=Sum of diagonal elements of A

Tr(B)=Sum of diagonal elements of B

C(Tr(A))= $C\cdot$ Sum of diagonal elements of A

C(Tr(B))= $C\cdot$ Sum of diagonal elements of B

$C(A+B)=C\cdot (A+B)$

Tr(C(A+B)=Sum of diagonal elements of (C(A+B))

Suppose ,A= $\left[\begin{array}{ccc}1&0\\1&1\end{array}\right]$

B= $\left[\begin{array}{ccc}1&1\\1&1\end{array}\right]$

Tr(A)=1+1=2

Tr(B)=1+1=2

C(Tr(A)+Tr(B))=C(2+2)=4C

A+B= $\left[\begin{array}{ccc}1&0\\1&1\end{array}\right]+\left[\begin{array}{ccc}1&1\\1&1\end{array}\right]$

A+B= $\left[\begin{array}{ccc}2&1\\2&2\end{array}\right]$

C(A+B)= $\left[\begin{array}{ccc}2C&C\\2C&2C\end{array}\right]$

Tr(C(A+B))=2C+2C=4C

Hence, Tr(C(A+B)=C(Tr(A)+Tr(B))

Hence, proved.

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3 years ago