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3 years ago

5

I need help with three math (algebra) problems?

1 answer:

Bumek [7]3 years ago

3 0

1.
A. (49-8)*2-16, 41*2-16, 82-16, 66, yep
B. 49-8*-14, 49+112, 161, nope
C. 7^(-16)-16, (1/(7^16))-16, nope
D. 49-16-16, 17, nope

A is answer

2.
2((a+b)²-b)
2((7+1)²-1)
2((8)²-1)
2(64-1)
2(63)
126
C is answer

4.
none of the options give the answer

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What is the solution of the system? Use elimination.

sweet [91]

Answer:

The solutions to the system of the equations by the elimination method will be:

$x=2,\:z=-1,\:y=2$

Step-by-step explanation:

Given the system of the equations

$2x\:+\:2y\:+z\:=\:7$

$-x-\:y\:+z\:=\:-5$

$x+3y-4z=12$

solving the system of the equations by the elimination method

$\begin{bmatrix}2x+2y+z=7\\ -x-y+z=-5\\ x+3y-4z=12\end{bmatrix}$

$\mathrm{Multiply\:}-x-y+z=-5\mathrm{\:by\:}2\:\mathrm{:}\:\quad \:-2x-2y+2z=-10$

$\begin{bmatrix}2x+2y+z=7\\ -2x-2y+2z=-10\\ x+3y-4z=12\end{bmatrix}$

$-2x-2y+2z=-10$

$+$

$\underline{2x+2y+z=7}$

$3z=-3$

$\begin{bmatrix}2x+2y+z=7\\ 3z=-3\\ x+3y-4z=12\end{bmatrix}$

$2x+6y-8z=24$

$-$

$\underline{2x+2y+z=7}$

$4y-9z=17$

$\begin{bmatrix}2x+2y+z=7\\ 3z=-3\\ 4y-9z=17\end{bmatrix}$

Rearranging the equations

$\begin{bmatrix}2x+2y+z=7\\ 4y-9z=17\\ 3z=-3\end{bmatrix}$

solve $3z=-3$ for z:

$z=-1$

$\mathrm{For\:}4y-9z=17\mathrm{\:plug\:in\:}z=-1$

solve $4y-9\left(-1\right)=17$ for y:

$4y-9\left(-1\right)=17$

$4y+9=17$

$4y=8$

$y=2$

$\mathrm{For\:}2x+2y+z=7\mathrm{\:plug\:in\:}z=-1,\:y=2$

solve $2x+2\cdot \:2-1=7$ for x:

$2x+2\cdot \:2-1=7$

$2x+3=7$

$2x=4$

$x=2$

Therefore, the solutions to the system of the equations by the elimination method will be:

$x=2,\:z=-1,\:y=2$

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3 years ago

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Step-by-step explanation:

The Euler's method is a common approximation method for the solution of differential equations. It can be used to obtain an approximate value of a given function. The Euler's formula is:

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$y_{1} = y_{0} + hF(x_{0},y_{0})$

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