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andrey2020 [161]

3 years ago

15

Number 18. ..........,

2 answers:

klio [65]3 years ago

8 0

Answer:

d to the power of 4

Step-by-step explanation:

ExtremeBDS [4]3 years ago

6 0

$\huge\textsf{Hey there!}$

$\text{Usually when you have a variable by itself it’s usually understood to be an INVISIBLE ONE}$

$\rm{d\timesd\times d \times d}$

$\rm{= 1d\times 1d \times1d\times1d}$

$\rm{= d^2 \times d^2}$

$\rm{= d^{2\times2}}$

$\rm{= d^4}$

$\huge\boxed{\mathsf{Therefore, \ your \ answer \ should \ be: d^4}}\huge\checkmark$

$\huge\text{Good luck on your assignment \& enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

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Solve the system of equation by guess sidle method

lilavasa [31]

Answer: The solution is,

$x_1\approx 0.876$

$x_2\approx 0.419$

$x_3\approx 0.574$

Step-by-step explanation:

Given equations are,

$8x_1 + x_2 + x_3 = 8$

$2x_1 + 4x_2 + x_3 = 4$

$x_1 + 3x_2 + 5x_3 = 5$ ,

From the above equations,

$x_1=\frac{1}{8}(8-x_2-x_3)$

$x_2=\frac{1}{4}(4-2x_1-x_3)$

$x_3=\frac{1}{5}(5-x_1-3x_2)$

First approximation,

$x_1(1)=\frac{1}{8}(8-(0)-(0))=1$

$x_2(1)=\frac{1}{4}(4-2(1)-(0))=0.5$

$x_3(1)=\frac{1}{5}(5-1-3(0.5))=0.5$

Second approximation,

$x_1(2)=\frac{1}{8}(8-(0.5)-(0.5))=0.875$

$x_2(2)=\frac{1}{4}(4-2(0.875)-(0.5))=0.4375$

$x_3(2)=\frac{1}{5}((0.875)-3(0.4375))=0.5625$

Third approximation,

$x_1(3)=\frac{1}{8}(8-(0.4375)-(0.5625))=0.875$

$x_2(3)=\frac{1}{4}(4-2(0.875)-(0.5625))=0.421875$

$x_3(3)=\frac{1}{5}(5-(0.875)-3(0.421875))=0.571875$

Fourth approximation,

$x_1(4)=\frac{1}{8}(8-(0.421875)-(0.571875))=0.875781$

$x_2(4)=\frac{1}{4}(4-2(0.875781)-(0.571875))=0.419141$

$x_3(4)=\frac{1}{5}(5-(0.875781)-3(0.419141))=0.573359$

Fifth approximation,

$x_1(5)=\frac{1}{8}(8-(0.419141)-(0.573359))=0.875938$

$x_2(5)=\frac{1}{4}(4-2(0.875938)-(0.573359))=0.418691$

$x_3(5)=\frac{1}{5}(5-(0.875938)-3(0.418691))=0.573598$

Hence, by the Gauss Seidel method the solution of the given system is,

$x_1\approx 0.876$

$x_2\approx 0.419$

$x_3\approx 0.574$

4 0

3 years ago

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4 years ago

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