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Natasha_Volkova [10]
3 years ago
12

I want help with this equation -1 + x = 8x - 8

Mathematics
2 answers:
Ilya [14]3 years ago
7 0

Answer:

x=1

Step-by-step explanation:

To solve for x, we need to get x alone on one side of the equation.

Add one to each side of the equation.  Subtract 8x from each side of the equation.  Divide both sides by -7.  You should get 1 as the answer.

-1+x=8x-8\\x=8x-7\\-7x=-7\\x=1

MAVERICK [17]3 years ago
6 0

Answer:

x = 1

Step-by-step explanation:

First combine like terms from both sides, in this case moving the -8 to the left side and the x to the right side.

You do this by doing the opposite operation for the values.

Add 8 to both sides and cancel out -8.

-1 + x = 8x -8

+8             +8

Subtract x from both sides and cancel out x.

7 + x = 8x

   -x     -x

Then it looks like this.

7 = 7x

Divide by 7 on both sides

x = 1

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3 years ago
Determine the equation of the line with slope −4 that passes through the point M(−2, 1).
scoray [572]
Y=mx+b
m=slope
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y=-4x+b 
Plug in the coordinate in place of x and y (-2,1) 
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b=-7

Final answer: 
<span>y = −4x − 7
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5 0
3 years ago
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Use the laplace transform to solve the given initial-value problem. y' 5y = e4t, y(0) = 2
Basile [38]

The Laplace transform of the given initial-value problem

y' 5y = e^{4t}, y(0) = 2 is  mathematically given as

y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}

<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>

Generally, the equation for the problem is  mathematically given as

&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}

\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}

put $s=-s \Rightarrow a_{1}=\frac{17}{9}$

\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}

In conclusion, Taking inverse Laplace tranoform

L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\

y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}

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6 0
2 years ago
HELP PLEASE!
pav-90 [236]

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6 0
3 years ago
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in the interval 0° ≤ x ≤360°, find the values of x for which tan x =1.3562 Give your answers to the nearest degree
belka [17]

Answer:

The answers are 54 degrees and 234 degrees

Step-by-step explanation:

mathematically we have the tan positive on only two axes

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We start by find the arc tan of the angle value

Thus;

x = arc tan (1.3562)

x = 54 degrees

On the third quadrant, we have it that;

180 + 54 = 234 degrees

4 0
3 years ago
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