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3 years ago

3x − 15= 3x + 15 what is x?

Mathematics

1 answer:

Marina CMI [18]3 years ago

7 0

We are given the following equation and want to find the possible values of x

$3x-15=3x+15$

Subtract both sides by 3x

$3x-15-3x=3x+15-3x$

$-15=15$

Now, we have this weird equation. Since -15 is NOT equal 15 and there is no other way we could have solved for x, there is no real solution to the given equation.

In other words, there is no value of x that makes the equation true.

Let me know if you need any clarifications, thanks!

~ Padoru

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What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

So,

$b=0,b-4=0$

$b=0,b=4$

Therefore, the solutions are:

$b=0,\:b=4$

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3 years ago

Read 2 more answers

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Mars2501 [29]

I think is inside and then outside. Not sure

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LenaWriter [7]

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So, from initial conditions we find resistivity(р)
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R*A /L=р
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<span>As, material remains same so resistivity(p) doesn't change

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Diameter= D1 = 5 cm= 5*10⁻² m
Radius= r= 0.025 m
Area= A1= π*r²/2 =9.8125*10⁻⁴ m²
R=р*L1/A1
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4 years ago