4 years ago

Is the set of all irrational real numbers countable?

Mathematics

1 answer:

MaRussiya [10]4 years ago

8 0

I think that they are.

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If f(x) = 5x^2– 3 and f(x + a) = 5x^2+ 30x + 42,<br> what is the value of a ?

Andrei [34K]

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The zeros are x=0,3,-2

There is a multiplicity of 1 for all of them.

Step-by-step explanation:

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3 years ago

Which angles are supplementary to 14 ? select all that apply.

Furkat [3]

Answer will be 15>
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Work out the size of angle ABC and give a reason for your answer

Mariulka [41]

Answer:

there is no picture so it is impossible to answer

Step-by-step explanation:

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3 years ago

PLEASE HELP ME!!!!!!!

Gennadij [26K]

-4x < 20
Divide -4 on both sides and the x would be by itself and 20/4 = 5. Since it is a negative, the sign would switch to >
This would be the inequality : x > 5

Numbers in the solution set would be any number greater than 5
ex: 6, 10, 15

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3 years ago

Log(x⁴+3x³) - log(X + 3 ) + log2 - log6 = 2logx . find the value of x

AfilCa [17]

The given equation is

$\begin{gathered} \log (x^4+3x^3)-\log (x+3)+\log 2-\log 6=2\log x \\ \log (\frac{x^4+3x^{3^{}}}{x+3})+\log \frac{2}{6}=\log x^2 \\ \log \frac{x^3(x+3)}{x+3}+\log \frac{1}{3}=\log x^2 \\ \log x^3+\log \frac{1}{3}=\log x^2 \\ \log \frac{x^3}{3}=\log x^2 \\ \frac{x^3}{3}=x^2 \\ x^3-3x^2=0 \\ x^2(x-3)=0 \end{gathered}$

hence

$x=0\text{ or x=3}$

But x cannot be zero so x=3

So the value of x is 3

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1 year ago