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3 years ago

How many solutions -(3 - 6b) = 6b + 5

Mathematics

1 answer:

Aleksandr [31]3 years ago

6 0

Answer:

There are no solutions

Steps :

First we expand it

$-3+6b=6b+5$

Then we add 3 to both sides

$-3+6b+3=6b+5+3$

Then we simplify

$6b=6b+8$

Then we subtract 6b from both sides

$6b-6b=6b+8-6b$

Simplify

$0=8$

No solutions !

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Y = -3x + 11 y = -6x - 13

Marysya12 [62]

Answer:

x= -8 and so y=35.....solving equation

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4 years ago

Determine the intercepts of the line. 4x-1=3y+5

guapka [62]

Calculate the intercepts bu replacing x or y with 0 and solving for the other.

4x - 1 = 3y + 5
If x = 0,
4(0) - 1 = 3y + 5
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-6 = 3y
y = -2

If y = 0
4x - 1 = 3(0) + 5
4x - 1 = 5
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x = 6/4 or 3/2

x int: (3/2,0)
y int: (0,-2)

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4 years ago

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GrogVix [38]

Take the root of both sides of the equation.
y = ^3 √64
Calculate the root.
y = 4

In the end, when you substitute y for 4,
4^3 will get you 64.

8 0

3 years ago

Read 2 more answers

Write the partial fraction decomposition fro the ration expression below

zlopas [31]

$\dfrac{5x-2}{(x-1)^2}=\dfrac a{x-1}+\dfrac b{(x-1)^2}$
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Since $5x-2=5x-5+3=5(x-1)+3$ , it follows that $a=5$ and $b=3$ , so

$\dfrac{5x-2}{(x-1)^2}=\dfrac5{x-1}+\dfrac3{(x-1)^2}$

$\dfrac{x^3+x^2+x+2}{x^4+x^2}=\dfrac{x^3+x+x+2}{x^2(x^2+1)}=\dfrac ax+\dfrac b{x^2}+\dfrac{cx+d}{x^2+1}$
$x^3+x^2+x+2=ax(x^2+1)+b(x^2+1)+(cx+d)x^2$
$x^3+x^2+x+2=(a+c)x^3+(b+d)x^2+ax+b$

When $x=0$ , you find that $b=2$ . It's also clear that $a=1$ . So the remaining constants must be $1+c=1\implies c=0$ and $2+d=1\implies d=-1$ , and so you get

$\dfrac{x^3+x^2+x+2}{x^4+x^2}=\dfrac1x+\dfrac2{x^2}-\dfrac1{x^2+1}$

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4 years ago

Which number is written in scientific notation? 6 times 10 Superscript 11 12 times 10 Superscript negative 8 0.2 times 10 Supers

Mandarinka [93]

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Step-by-step explanation:

Scientific notation is in the form ;

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